GIFT  OF 
the  estate  of 

Professor  William  F.  Mever 


asm 


A  SHOET  COURSE  ON 
DIFFERENTIAL  EQUATIONS 


t. 


A  SHORT  COURSE  ON 


DIFFERENTIAL  EQUATIONS 


BY 


DONALD   FRANCIS   CAMPBELL,  PH.D. 

PROFESSOR  OF  MATHEMATICS,   ARMOUR  INSTITUTE  OF  TECHNOLOGY 


Nrto  ¥orfe 

THE   MACMILLAN   COMPANY 

LONDON:  MACM1LLAN  &  CO.,  LTD. 

1907 

Alt  rights  reserved 


ASTRONOMY  LIBRARY 

COPYRIGHT,  1906 
BY  THE   MACMILLAN   COMPANY 


Set  up  and  electrotyped.     Published  September,  1906 
Enlarged  October,  1907. 


PRESS  OP 

THC  NEW  ERA  PRINTING  COMPANY 
LANCASTER,  PA.    • 


ory 


PREFACE 


In  many  Colleges  of  Engineering,  the  need  is  felt  for  a  text- 
book on  Differential  Equations,  limited  in  scope  yet  comprehen- 
sive enough  to  furnish  the  student  of  engineering  with  sufficient 
information  to  enable  him  to  deal  intelligently  with  any  differen- 
tial equation  which  he  is  likely  to  encounter.  To  meet  this  need 
is  the  object  of  this  book. 

Throughout  the  book,  I  have  endeavored  to  confine  myself 
strictly  to  those  principles  which  are  of  interest  to  the  student  of 
engineering.  In  the  selection  of  problems,  the  aim  was  con- 
stantly before  me  to  choose  only  those  that  illustrate  differential 
equations  or  mathematical  principles  which  the  engineer  may 
meet  in  the  practice  of  his  profession. 

I  have  consulted  freely  the  Treatises  on  Differential  Equations 
of  Boole,  Forsyth,  Johnson,  and  Murray.  I  am  indebted  to  two 
of  my  colleagues,  Professors  N.  C.  Riggs  and  C.  W.  Leigh,  for 
reading  parts  of  the  manuscript  and  verifying  many  of  the 

answers  to  problems. 

D.  F.  CAMPBELL. 
CHICAGO,  ILL., 
September,  1906. 


PREFACE  TO  ENLARGED  EDITION 


This  book  as  it  first  appeared  consisted  of  the  first  eight 
chapters  as  here  given.  The  kindly  criticism  by  a  number  of 
those  teachers  for  whose  use  it  was  intended  on  the  need  of  a 
discussion  of  equations,  that  occur  in  investigations  in  Mathe- 
matical Physics,  other  than  those  given  in  these  chapters  has 
induced  me  to  add  Chapter  IX  to  the  book. 

M577086 


VI  PREFACE 

In  the  preparation  of  Chapter  IX.,  I  have  drawn  freely  from 
Professor  Byerly's  Treatise  on  Fourier's  Series  and  Spherical 
Harmonics,  from  Professor  Bocher's  pamphlet  entitled  Regular 
Points  of  Linear  Differential  Equations  of  the  Second  Order  and 
from  notes  kindly  loaned  me  by  Professor  Snyder  of  Cornell 
University.  I  have  also  consulted  Heffter's  Treatise  on  Linear 
Differential  Equations  with  one  Independent  Variable. 

To  those  teachers  who  have  sent  me  their  criticism  of  the 
book  in  its  original  form,  as  well  as  to  others  who  have  cordially 
received  it,  I  am  under  the  deepest  obligations. 

D.  F.  CAMPBELL. 
CHICAGO,  ILL., 

June,  1907. 


CONTENTS 

PAGES 
CHAPTEE  I 

INTRODUCTION 1-15 

CHAPTER  II 
CHANGE  OF  VARIABLE 16-21 

CHAPTER  III 

ORDINARY  DIFFERENTIAL  EQUATIONS  OF  THE  FIRST  ORDER  AND 

FIRST  DEGREE 22-40 

CHAPTER  IV 

ORDINARY  LINEAR  DIFFERENTIAL  EQUATIONS  WITH  CONSTANT 

COEFFICIENTS 41-64 

CHAPTER  V 

HOMOGENEOUS   LINEAR   DIFFERENTIAL    EQUATIONS.      EXACT 

LINEAR  DIFFERENTIAL  EQUATIONS 65-70 

CHAPTER  VI 
CERTAIN  PARTICULAR  FORMS  OF  EQUATIONS    .       ,        .        .      71-76 

CHAPTER  VII 
ORDINARY   DIFFERENTIAL   EQUATIONS   IN   Two   DEPENDENT 

VARIABLES ...        .        .      77-88 

CHAPTER  VIII 
PARTIAL  DIFFERENTIAL  EQUATIONS  .        .       ; .  •    ..        .        .      89-96 

CHAPTER  IX 

APPLICATIONS  OF  PARTIAL  DIFFERENTIAL  EQUATIONS.    INTE- 
GRATION IN  SERIES 97-123 


yii 


A  SHORT  COURSE 

*      ON 

DIFFERENTIAL  EQUATIONS 


CHAPTER  I 
INTRODUCTION 

1.  There  are  various  definitions  given  for  a  function  of  one 
variable.  We  shall  here  adopt  the  following  : 

If  to  every  value  of  x  there  corresponds  one  or  more  values  of 
/(#),  then  /(#)  is  said  to  be  a  function  of  x. 

This  definition  includes  a  constant  as  a  function  of  x,  for  if 
/(#)  is  constant,  then  for  every  value  of  x,  /(#)  has  a  value, 
namely,  this  constant. 

A  definition  of  a  function  of  two  variables  is  the  following  : 

If  to  every  pair  of  values  of  two  variables  x  and  y  there  cor- 
responds one  or  more  values  of/(#,  ?/),  then/(#,  ^)  is  said  to  be 
a  function  of  x  and  y. 

This  includes  a  constant  or  a  function  of  one  variable  as  a 
function  of  x  and  y.  .  ; 

A  function  /(#)  of  one  variable  x  is  single  valued  when  for 
every  value  of  x  there  is  one  and  only  one  corresponding  value 


A  function  /(x*)  of  one  variable  x  is  continuous  for  a  value 
x  =  a  if  /(a)  is  finite,  and 


A  function  f(x,  y)  of  two  independent  variables  x  and  y  is 
single  valued  when  for  every  set  of  values  for  x  and  y  there  is 
one  and  only  one  corresponding  value  of/(#,  y). 

1 


2       SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

A  function  f(x,  y*)  of  two  independent  variables  x  and  y  is 
continuous  for  a  set  of  values  x  =  a,  y  =  b  if  /(a,  6)  is  finite, 

and 

limit 


no  matter  how  h  and  k  approach  zero. 

The  following  definitions  are  given  in  almost  any  work  in 
calculus  : 

If  /(#)  is  a  single  valued  and  continuous  function  of  x,  given 
by  the  equation  y  =  /(#),  then 

Ao;  and  A?/  denote  the  increments  of  x  and  y  respectively, 

<fy        limit 


dy  =  -^  dx. 
dx 

If  /(#)  is  single  valued  and  continuous,  and  dy/dx  is  contin- 
uous, then 

dx2  ~  dx\dx 

In  general,  if  /(#)  is  single  valued  and  continuous,  and  the 
preceding  derivatives  are  all  continuous,  then 


dx"  ~"  dx  \  dxn 

If  /(#,  y)  is  a  single  valued  and  continuous  function  of  two 
independent  variables  x  and  y,  given  by  the  equation  z  =/(#,  t/), 
then  dz/dx  is  the  derivative  of  2  with  respect  to  x  when  y  is  held 
constant ;  dz/dy  is  the  derivative  of  z  with  respect  to  y  when  x  is 
held  constant. 

2.  In  a  single  valued  and  continuous  function  /(#)  of  one  vari- 
able x,  given  by  the  equation  y  =/(#),  whether  x  is  the  inde- 
pendent variable  or  a  function  of  some  other  variable  or  variables, 
we  have 


INTRODUCTION 
=  d(dx)  ;  fx  =  d(d*x~)  ;  •  •  •  ;  dnx  = 


Definitions.  The  differentials  cfo,  cPrc,  cf#,  •  •  •,  dnxy  or 
dy,  d2y,  d*y,  •  •  •,  dni/  are  called  the  first,  second,  third,  •  •  •,  nth, 
differentials  respectively. 

3.  Derivation  of  d2y  and  cPy  when  no  assumption  is  made  re- 
garding x  being  independent  or  a  function  of  some  variable  or 
variables. 


dx 

By  taking  differentials  in  succession  any  differential  may  ulti- 
mately be  found. 

4.  In  the  differentials  of  the  preceding  article,  if  x  is  an  inde- 
pendent variable,  it  can  be  assumed  without  loss  of  generality, 
that  Arc,  or  what  is  the  same  in  this  case,  dx,  is  constant.  That 
is,  it  can  be  assumed  that  x  changes  by  equal  increments.  Under 
this  supposition,  therefore,  d*x  and  all  higher  differentials  of  re 
can  be  taken  zero.  Therefore,  under  this  supposition, 


The  place  which  a  derivative  or  differential  occupies  in  the 
succession  of  derivatives  or  differentials  indicates  the  order  of  the 
derivative  or  differential.  Thus,  a  second  derivative  or  differ- 
ential is  said  to  be  of  the  second  order,  a  third  of  the  third  order, 
and  so  on. 


4       SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

5.  The  only  functions  usually  considered  in  elementary  works 
in  calculus  are  functions  of  a  real  variable.  Such  functions  with 
one  exception  are  the  only  ones  considered  in  the  following  pages. 
The  exception  is  tf  where  2  is  a  complex  quantity. 

The  student  is  already  familiar  with  the  definition  of  ex  where 
x  is  real.  He  is,  however,  probably  not  familiar  with  the  defini- 
tion of  e*  when  z  is  a  complex  quantity.  A  definition  of  this 
function  will  now  be  given. 

The  infinite  series 

z2        s8 


where  z  is  a  complex  quantity,  can  be  shown  to  have  a  determi- 
nate, finite  value  for  every  value  of  z.  It  also  reduces  to  the 
infinite  series 

'  of       a? 


when  z  becomes  real  and  equal  to  x,  and  this  series,  it  will  be  re- 
membered, is  equal  to  ex  for  all  values  of  x.  It  therefore  appears 
that  the  infinite  series  in  z  would  be  satisfactory  as  a  definition 
of  ef.  We  shall  define  e?  by  saying  that  it  is  equal  to  the  infinite 
series 


for  all  values  of  z. 

From  this  definition,  the  following  theorem  can  be  established  : 
Theorem.     If  z  =  x+yj  where  x  and  y  are  real,  and  j  =  V—  1, 

then 

e*  = 


>  l>7  Definition. 


Proof. 


1    L  f*      ^  _i_  O  +  20')*  _,_  O  +  3P')8  , 
=  1  +  O  +  W)  +  --  12  --  +  --  [3  --  + 


INTRODUCTION  5 

Consider  all  the  terms  containing  of.     These  are  found  from  the 
terms 

O  +  w?     O  +  yJY+1 


They  are 


|r  |r  +  l 


or 


Separate  the  real  terms  from  the  imaginary  and  there  results, 


or 

xr 


Let  r  take  all  positive  integral  values  in  succession  from  0. 
In  this  way  we  get  all  the  terms  of  the  development  e?.     Then 


=  I 


< 
I  +  x  +  —  + 


j  sin  y). 

The  theorem  is  therefore  proved. 
EXAMPLES.     e~x+Za!f  =  e~*(cos  Bx 

EXERCISES 

1.  Given  y  ==  log  x,  find  %,  dfy,  d*y  : 

(a),  on  the  assumption  that  x  is  the  independent  variable  ; 

(6),  making  no  assumption  with  regard  to  x. 

In  the  results  of  (6),  substitute  x  =  cos  0  and  show  that  the 
results  are  the  same  as  those  obtained  by  first  substituting  the 
value  of  x  in  log  x  and  then  taking  the  differentials. 

2.  Given  y  =  ex  where  x  =  cos  9,  express  dy,  d2y,  d?y  in  terms 
of  0  without  substituting  the  value  of  x  in  the  equation  y  =  ec. 


6        SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

3.  Given  y  =  log  x  where  x  =  sin  0,  express  dy,  d*y,  d*y  in 
terms  of  0  without  substituting  the  value  of  x  in  the  equation 
y  =  log  x. 

4.  Prove  that  e*"*  =  e*(cos  y  —  j  sin  y). 

5.  Prove  that  &+*<!***  =  e*+z+(y+wv,  where  x,  y,  z  and  w  are 
real. 

ANSWERS 


.  =  _  =  _^-  —  p. 

/t\     j       ^     j«  **  * 

(6).   rfy  =-  ;  cfy= 


2.  dy  =  —  smO  eco*'dO  ; 
d*y=—  sivO  ecosed*0  +  (sin'fl 
cPy  =  —  sin0  ecos^30  +  3(sin20  — 

+  sin0(l  +  3  cos0  - 

3.  dy  =  cot0  e^0  ;  cPy  =  cot0  c?20  —  cosec20  c?02  ; 

f  y  =  cotO  (PO  —  3  cosec20  d0d20  +  2  cosec20  cot0  c?08. 

6.  Definition  of  differential  equation.  A  differential  equation 
is  an  equation  involving  derivatives  or  differentials  with  or  with- 
out the  variables  from  which  these  derivatives  or  differentials  are 
derived. 

The  following  are  examples  of  differential  equations  : 


dx 


m 


(3) 


dx 


INTRODUCTION 


7.  In  examples  (1)  to  (4)  inclusive  of  the  preceding  article 
it  will  be  noticed  that  differentials  enter  the  equation  only  in  de- 
rivatives. It  is  conceivable,  however,  that  there  might  be  an 
equation  containing  differentials  other  than  those  in  the  deriva- 
tives, as  for  example, 


but  there  is  no  need  of  entering  into  a  discussion  of  such  equa- 
tions, and  we  shall  not  do  so.  In  what  follows,  we  shall  assume 
that  if  the  equation  is  written  in  differential  form,  the  differen- 
tials can  all  be  converted  into  derivatives  by  the  process  of 
division. 

8.  Classes  of  differential  equations.     Differential  equations 
are  divided  into  two  classes  :  ordinary  and  partial. 

An  ordinary  differential  equation  is  one  in  which  all  the 
derivatives  involved  have  reference  to  a  single  independent 
variable. 

A  partial  differential  equation  is  one  which  contains  partial 
derivatives  and  therefore  indicates  the  existence  of  two  or  more 
independent  variables  with  respect  to  which  these  derivatives 
have  been  formed. 

Thus,  in  Art.  6,  equations  (1),  (2),  (3)  and  (4)  are  ordi- 
nary differential  equations,  and  equations  (5)  and  (6)  are  par- 
tial differential  equations. 

Chapters  I  to  VII  inclusive  are  devoted  to  a  discussion  of  ordi- 
nary differential  equations.  Chapter  VIII  contains  a  short 
treatment  of  some  partial  differential  equations. 

9.  Order  and  degree  of  a  differential  equation.    The  order  of 
a  differential  equation  is  that  of  the  highest  derivative  or  differ- 
ential in  the  equation. 


8       SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

Thus,  in  Art.  6,  equations  (1)  and  (4)  are  of  the  third  order, 
and  (2)  and  (3)  of  the  second  order. 

The  degree  of  a  differential  equation  is  the  degree  of  the  deriv- 
ative or  differential  of  highest  order  in  the  equation  after  the 
equation  is  freed  from  radicals  and  fractions  in  its  derivatives. 

Thus,  in  Art.  6,  equation  (1)  is  of  the  second  degree,  equa- 
tions (2),  (3)  and  (4)  of  the  first  degree. 

10.  Solutions  of  a  differential  equation.  Let  us  consider  the 
differential  equation  in  each  of  the  two  following  examples,  and 
see  if,  from  the  equation,  we  can  get  a  relation  connecting  x  and 
y  and  not  involving  derivatives,  such  that,  if  the  value  of  y  in 
terms  of  x  be  substituted  in  the  equation,  the  equation  is  satisfied. 


EXAMPLE  1. 
By  integration,  we  get 


-    =  x2. 
ax 


EXAMPLE  2.  -y4  +  y  =  0. 

Multiply  the  equation  by  2dy/dx  and  integrate. 


In  example  1,  if  ^ 
there  results  x2  =  x2. 
In  example  2,  if 


.- 
dx 


dy 


±dx. 


.  y  =  ±  Vc  sin  (x  -f 


or    y  =  rb 


(#  -f-  c2). 


*  +  c  be  substituted  for  y  in  the  equation, 
The  equation  is  therefore  satisfied. 
Vc  sin  (x  -f-  c,),   or  ±  Vc  cos  (x  -f-  c2)  be 


substituted  for  y  in  the  equation,  there  results,  in  the  first  case, 


INTRODUCTION  9 

=P  Vc  sin  (x  -f  Cj)  =b  Vc  sin  (x  -f  cj  =  0,  and  in  the  second  case, 
ip  Vc  cos  (#  -f-  c2)  =b  Vc  cos  (x  4-  c2)  =  0.  In  either  case  the 
equation  is  satisfied. 

Definition.  A  solution  of  a  differential  equation  is  a  relation 
between  the  variables  of  the  equation  and  not  involving  deriva- 
tives, such  that  if  the  value  of  the  dependent  variable  be  substi- 
tuted in  the  equation,  the  equation  is  satisfied. 

Thus,  y  =  J#3  4-  c  of  example  1,  and  y  =  ±  Vc  sin  (x  -j-  Cj) 
of  example  2,  are  solutions  of  the  equations. 

In  this  book  we  shall  not  concern  ourselves  with  the  question 
of  whether  every  differential  equation  has  a  solution  but  shall  be 
content  with  finding  solutions  in  the  few  special  cases  discussed 
here. 

11.  A  solution  of  an  ordinary  differential  equation  may  be 
one  of  three  kinds:  general,  particular  and  singular. 

A  general  solution  is  one  which  contains  arbitrary  constants 
equal  in  number  to  the  exponent  of  the  order  of  the  equation. 

Thus,  in  example  1,  Art.  10,  the  number  of  arbitrary  con- 
stants is  one  and  the  exponent  of  the  order  of  the  equation  is  1, 
and  in  example  2  of  the  same  article  the  number  of  arbitrary 
constants  is  two,  and  the  exponent  of  the  order  of  the  equation 
is  2.  In  either  case  the  solution  is  the  general  solution  of  the 
equation. 

A  particular  solution  of  a  differential  equation  is  a  solution 
obtained  from  the  general  solution  by  giving  one  or  more  of  the 
constants  particular  values. 

Thus 

af  x*  x* 

y  =  ~>     y  =-o  +  l     or    y  =  -  _  5, 

'    o  o  o 

of  example  1,  Art.  10,  or  y  =  sin  x,  y  =  2smx,  or  y=  —  3  cos  x, 
of  example  2  of  the  same  article,  are  particular  solutions  of  the 
equations. 

A  singular  solution  of  a  differential  equation  is  a  solution  with- 
out arbitrary  constants  which  cannot  be  derived  from  the  general 
solution  by  giving  the  constants  particular  values. 

2 


10      SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

Singular  solutions  will  not  be  considered  in  this  book. 

12.  A  solution  of  a  differential  equation  is  not  a  general  solu- 
tion unless  the  constants  are  in  number  equal  to  the  exponent  of 
the  order  of  the  equation,  and  cannot  be  reduced  to  a  fewer 
number  of  equivalent  constants. 

Thus,  y  =  ce*+a,  c  and  a  arbitrary  constants,  although  it  con- 
tains two  arbitrary  constants,  is  not  the  general  solution  of  a  dif- 
ferential equation  of  the  second  order,  as  can  readily  be  shown. 
The  equation  y  =  ce?+a  is  the  same  as  y  =  cV.  Now  ca  is  equiv- 
alent to  only  one  arbitrary  constant  because  an  arbitrary  con- 
stant can  have  any  value  and  thus  all  the  particular  solutions 
got  by  giving  c  and  a  all  possible  values  can  be  obtained.  There- 
fore y  =  ca€?  is  equivalent  to  a  solution  y  =  Ae*,  A  arbitrary,  and 
cannot  therefore  be  the  general  solution  of  a  differential  equation 
of  the  second  order. 

13.  Let  y  =/x(a?),  y  =/2O),  •  •  •,  y  =  /n(a?)  be  solutions  of  a 
differential  equation. 

Definition.  If  the  c's  cannot  be  chosen,  not  all  zero,  such  that 
cifi(x)  +  c2/a(^)  +  " ' '  +  cnfn(x)  ig  identically  zero,  then  the 
solutions  are  said  to  be  linearly  independent. 

Thus,  y  =  ±  Vc  sin  (x  +  Cj)  and  y=  =t  Vc  cos  (x  +  c2)  of  ex- 
ample 2,  Art.  10,  are  such  that  no  values  cs  and  c4,  not  both 
zero,  can  be  chosen  such  that  ±  cs  Vc  sin  (a+Cj)  ±c4  Vc  cos  (#+ca) 
is  identically  zero.  The  solutions  are  therefore  linearly  inde- 
pendent. 

14.  Derivation  of  an  ordinary  differential  equation.     Let 

<K*,  y,  O  =  0  (1) 

be  an  equation  containing  x  and  y,  and  the  arbitrary  constant  cr 
By  differentiation  of  (1)  there  results 

d$      d^dy 
d~x+dydx  =  Q' 

Equation  (2)  will  in  general  contain  cr     If  between  (1)  and 


INTRODUCTION  11 

(2),  Cj  be  eliminated,  the  result  is  a  differential  equation  of  the 
first  order  of  which  <j>(x,  y,  cx)  =  0  is  the  general  solution. 

EXAMPLE.     Find  the  differential  equation  of  which 

m 

y=2+Cl6 
is  the  general  solution. 

^-       2cxe~*2 
dx  -          l 

Eliminate  cx  between  the  equations.     Therefore 

-^  4-  2xy  =  mx 
dx^ 

is  the  differential  equation  of  which 

m 

y  =  2  +  v 

is  the  general  solution. 

Sometimes  the  arbitrary  constant  is  so  involved  that  it  disap- 
pears in  the  equation  which  results  from  the  differentiation.  In 
such  a  case  this  equation  is  the  desired  equation. 

EXAMPLE.     Find  the  differential  equation  of  which  y*  = 
is  the  general  solution. 

Divide  both  sides  of  the  equation  by  x. 


By  differentiation  there  results 


which  is  the  desired  differential  equation. 

Let 

+(x,  y,  cv  c2)  =  0  (1) 

be  an  equation  between  x  and  y,  and  two  arbitrary  constants  cl 
and  cr 


12      SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 
By  differentiation  of  (1)  there  results 

d<f>      a*  dy 

dx  +  dy  dx  = 

Equation  (2)  contains  dy/dx  and  will  in  general  contain  cx  and 
c2  also.  Eliminate  one  of  the  constants  between  the  two  equa- 
tions. Suppose  the  constant  cx  to  be  eliminated.  The  resulting 
equation  contains  dy/dx  and  in  general  x,  y  and  c2.  Call  it 


By  differentiation  there  results 


r          =  - 

ox  '  dydx 

Equation  (3)  contains  d2y/dx*  and  will  in  general  contain  c2. 
Eliminate  c2  between  (2)  and  (3).  The  result  is  a  differential 
equation  of  which  <f>(x,  y,  cv  c2)  =  0  is  the  general  solution. 

EXAMPLE.     Find  the  differential  equation  of  which 


is  the  general  solution. 

/» 

Differentiate  y  =  ^x  +  -?. 

x 


.        _  , 

•  dx  -   l  ~  if 
Eliminate  c,. 

.  «  xdy  -^ 

•  •  y  —  »c  ~T~  ==  —  • 
dx       x 

Differentiate. 

.  *y  **§ 

•  dx*~  x*' 
Eliminate  c2  between 

&y    2c2      j          ^y  _  2c2 

dx3  ~  x3  dx  ~~  x 


INTRODUCTION  13 


..  = 

ax2         dx 
which  is  the  desired  differential  equation. 

15.  It  is  seen  from  the  preceding  article  that  one  constant  can 
be  removed  after  each  differentiation.     From  this  it  would  be 
expected  that,  starting  with  the  differential  equation,  an  arbi- 
trary constant  might  be  introduced  every  time  the  order  of  the 
differential  equation  was  lowered  by  unity.     Then,  since  lower- 
ing the  order  of  a  differential  equation  of  the  nth  order  by  unity 
n  times  would  result  in  a  solution  of  the  differential  equation,  it 
would  be  expected  that  a  solution  would  contain  not  more  than 
n  arbitrary  constants. 

It  is  a  theorem  that  a  differential  equation  cannot  contain  a 
solution  having  more  arbitrary  constants  than  the  exponent  of 
the  order  of  the  equation  unless  the  constants  are  such  that  they 
can  be  reduced  to  a  fewer  number  of  equivalent  constants.  This 
will  be  assumed  without  further  discussion. 

It  is  also  a  theorem  that  a  differential  equation  cannot  have 
more  than  one  general  solution.  This  theorem  will  be  assumed 
without  discussion. 

16.  A  general  solution  may  have  various  forms  but  there  is 
always  a  relation  between  the  constants  of  one  form  and  those  of 
another.     Thus,  the  general  solution  of  example  2,  Art.  10,  may 
be  written  y  =  A  sin  x  -f  B  cos  x  instead  of  y=  ±  Vc  sin  (z-fCj). 
This  latter  form  of  solution  is  t/  =  ±.  Vc  cos  cx  sin  x±  V^sin  ^  cos  x, 
so  that  A  =  ±  V0  cos  cv  and  B  =  ±  V^sin  cr 


EXERCISES 
Determine  the  order  and  degree  of  the  following  equations. 


14       SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

In  each  of  the  seven  following  exercises  determine  the  differ- 
ential equation  of  which  the  given  equation  is  the  general  solu- 
tion, given  that  cl9  c2  and  c  are  arbitrary  constants. 

2.  y  =  GI  sin  mx  -f-  c2  cos  mx.  5.  y  =  ex  -\-  c  _  c3. 

*    3.  v  =  ct  cos  (mt  -j-  c2).  ^    6.  xy  =  c^e*  -f  c2e~x. 

r"   4.   (a?  —  Cj)2  +  (y  —  c2)2  =  m2.    r    7.  y*  —  2ca?  —  c2  =  0. 


9.  Show  that 
is  a  solution  of 
v     10.  Show  that 


3  d2y      A 
_  _  2  _  o 

«  dx* 


5 

^  =  3^  +  Cl 
is  a  solution  of 


11.  Show  that 

y  = 
is  a  solution  of 


12;  Show  that 


r    ' 
is  a  solution  of 

c^2v      Z  dv 


ANSWERS 


INTRODUCTION  15 


CHAPTER  II 
CHANGE  OF  VARIABLE 

17.  Interchange  of  Variables.  It  is  sometimes  desirable  to 
transform  an  expression  involving  derivatives  of  the  function  y, 
in  y  =  /(#)  where  x  is  the  independent  variable,  into  an  equiva- 
lent expression  involving  derivatives  of  the  function  x,  given  by* 
the  same  equation,  where  y  is  the  independent  variable. 

The  formulas  for  such  a  transformation  can  be  readily  estab- 
lished as  follows: 


limit 


limit 


(i) 


= 

dx2  ~~  dxdx 


_ d.(fy\  ty 

~~  dy\dx/     dx9 

.^/l\.l 

dyl  dx  J     dx9 
\dy/     dy 


df 


d*x 


± 

dx 
dy  /      dy 


dz      dz     dy 

since  j-  =  j-  •  -j -. 

dx      dy     dx 

by  substitution  from  (1). 


(2) 


(fa; 


dx9      dx\dx*)       dy\        /^?V' 

\dy)  J 


_ 

dx 
dy 


16 


CHANGE  OF  VARIABLE  17 


/dx\*d?x         /d?x\2  /dx\*  —  — 

'(dy)  dy*  +  6(df)   (dy)         ~  dy*  dy 


dx  /dx\& 


/ 

\ 


Ty 


The  method  of  procedure  for  higher  derivatives  is  evident. 
The  transformations  to  which  these  formulas  apply  are  called 
change  of  the  independent  variable  or  interchange  of  variables. 

EXAMPLE.     Change  the  independent  variable  from  x  to  y  in 
the  equation 


™y  ™  y  y  I     y  i  o 

dxdx*~  dx2(dx;  '' 

Substitute  from  (1),  (2)  and  (3). 

d*x    ] 2  d?x  dx          ,  ^  „ 

~~T~2  i  —   ~T~\   ~T~    4-  O   I    ~T~5 

3^-^ 


(dxVl       dx 
\dy)  J       dy 


dy 

+ 


fdx 

I*     -  •  -      'dy> 

I <P a 

\dyt 


dx 


18.  Change  of  the  dependent  variable.  Suppose  that  y  is  a 
function  of  x  and  at  the  same  time  is  a  function  of  some  other 
variable  z.  The  derivatives  of  y  with  respect  to  x  can  then  be 
expressed  in  terms  of  derivatives  of  z  with  respect  to  x. 

As  a  function  of  z,  let  y  =  <t>(z).  Denote  differentiation  with 
respect  to  z  by  primes.  Then 

dy      dydz       d<j>(z)  dz  dz 

dx  ~~  dz  dx  ~      dz     dx~  dx' 

d*y       d  /dy\      d<j>'(z}  dz   t  ^  f*      ^;  ,  /dz\ 2  §       ,  ,  J20 


18     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 
eft/       d  (d*y\         ,„,  NAfe\s      Oj/r,  .dztfz  .  efz 

a?  -  d*  (<£)  =  *  (s)  U)  +  *  (*}  s©  +  *  (0)  a? 


Similarly  for  higher  derivatives. 

The  above  transformation  is  called  change  of  the  dependent 
variable. 

EXAMPLE.     In  the  equation 


change  the  dependent  variable  from  y  to  z  where  y  =  tan  z. 

dy  dz 

-f-  =  see's-,-. 
dx  dx 

d?y      0      „  (dz 

-JT^  =  2  sec  z  tan  z  I  -^- 
dXi  \  dx 

Substitute  in  the  equation. 

/^z\*          L    ^      /o  x  -i\       t   /^2\2 

.*.  2  sec  z  tan  2!  -=-  )  -j-sec  0-^  —(2  tans— 1)  sec  21-7-  ) 
\ax/  ax*  \dx/ 


dz\*     ^    dz 


3a;  sec4  z^  =  0. 
dx 


19.  Change  of  the  independent  variable.  Suppose  that  y  is 
a  function  of  x  where  #  is  a  function  of  some  other  variable  z. 
The  derivatives  of  y  with  respect  to  x  can  then  be  expressed  in 
terms  of  derivatives  of  y  with  respect  to  z. 

As  a  function  of  z  let  x  =  <^(z).  Denote  differentiation  with 
respect  to  z  by  primes.  Then 

dy      dydz      dy  1        dy     I 
dx  ~~  cfeda?  ~~  cfed#  ~  dz$'(z)' 
~ 


CHANGE  OF  VARIABLE  19 


.-.  __ 

da?  ~  dx\do?     ~  dz     {4>'(z)}2  dz*  ~  {<£'(«)}3  dz    dx 


{^00  }6  <b. 

Similarly  for  higher  derivatives. 

The  above  transformation  is  called  change  of  the  independent 
variable. 

EXAMPLE.     In  the  equation 

cfy  a?      efy  y 

S?~"l_a*S"hl  -aj2- 

change  the  independent  variable  from  x  to  2,  where  x  =  cos  2. 

dy      dy  dz  dy 

~^    ^^   i    ~^    is  —  cosec  3  7  • 
aa?       as  aa;  02 

ci2y       d  (dy\        d  /  dy\l 

~  =  -T-  (  /  )  =  -7-  I  -  cosec  2-/  )  -j- 
dar      dx\dx/       dz  \  dz/dx 

dz 
=  —  cosec2  2  cot  2  -^  4-  cosec2  2  ~. 

C?2  T  C?22 

Substitute  in  the  equation. 

.  '  .  cosec2  2  -^  —  cosec2  2  cot  2  -^  +  cosec2  2  cot  2  -^+cosec22  y  =  0. 


When  changing  either  the  dependent  or  independent  variable 
to  a  third  variable,  it  is  better  to  work  out  each  derivative  in  the 
particular  case  considered  rather  than  use  the  derivatives  ex- 
pressed in  the  general  case  as  formulas. 


20     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

•  ^,  EXERCISES 

In  each  of  the  four  following  exercises,  change  the  independent 
variable  from  x  to  y. 


dx 
2-|i 

=  p. 


/    4  w  v  ^  „ 

dx*  dx       "  \  dx2 )       "  c?a;2  \dx)  "    ^  ^da;/  "    *"  \dx 

In  each  of  the  two  following  exercises  change  the  dependent 
variable  from  y  to  z. 


where  y  =  tan  2. 


where  y  =  e*. 

In  each  of  the  four  following  exercises,  change  the  independent 
variable  from  re  to  2. 

7.  x*  -r^!  -f  #  -7-  +  v  =  0,  where  x  =  ef. 

dzr        dx 

8.  (1  —  a?)  -j4  —  #  ~jj-  =  0,  where  ^  =  sin  z. 

9.  z3  j-*  +  2aj*  j-=  +  v  =  0,  where  #  =  e*. 


cv  a;     cv  v  A      1 

10.    j-i  +  =  --  2  T"  +  71  --  2N2  =  °>  where  x  =  tan  z. 
dc1   •    1  +  x2dx  ^  (1  +  a;2)2 


CHANGE  OF  VARIABLE  21 

11.  Transform  the  formula  for  radius  of  curvature, 


p  = 


dx* 


into  polar  coordinates,   the  equations  of  transformation  being 
x  =  r  cos  0,  y  =  r  sin  0. 


ANSWERS 
2  d2x       dx 


^  n 

3.  p  =  —  *  -  T2-^-^--     4.   j-i  +  3  j^  +  2  3-  +  ^  =  0. 

*  2  T 


~z  »  rt   c25 

5.  T-i  —  2  T-  =  sm  2-2       6-  ^-§  —  2rc:o 
dar        ^      *—    /  c?a;3         or 


-.A  A 

10.   j-j  +  v  =  0. 
da1  T 


CHAPTER  III 

ORDINARY  DIFFERENTIAL  EQUATIONS  OF  THE  FIRST  ORDER 
AND  FIRST  DEGREE 

20.  An  ordinary  differential  equation  in  one  dependent  vari- 
able, of  the  first  order  and  first  degree,  may  be  represented  by 
the  equation 

Mdx  +  Ndy  =  0 

where  M  and  N  are  functions  of  x  and  y  and  do  not  contain 
derivatives. 

The  equation  Mdx  +  Ndy  =  0  cannot  be  integrated  in  the 
general  form.  There  are  certain  particular  forms  of  it,  however, 
which  can  be  integrated.  Some  of  these  will  now  be  investigated. 

21.  LINEAR  DIFFERENTIAL  EQUATIONS  OF  THE  FIRST  ORDER 
Definition.     An  ordinary  linear  differential  equation  of  the 

first  order  is  an  equation  in  the  form 


where  P  and  Q  are  functions  of  x  and  do  not  contain  y  or  deriv- 
atives. • 

The  general  solution  of  the  equation 


can  be  found  as  follows  : 

Multiply  both  sides  of  the  equation  by  ef  pdx. 


22 


EQUATIONS  OF  FIEST  ORDER  AND  FIRST  DEGREE  23 

If  the  substitution  u  =  yeJ  pdx  be  made,  the  left  hand  member  of 
the  equation  reduces  to  du/dx. 


=  fQeSPdxdx 


.  •  .  y  =  e-SpdxfQefPdxdx  +  ce-f™,  (1) 

which  is  the  general  solution  of  the  equation. 

In  the  original  equation,  if  P  is  zero,  the  equation  reduces  to 
the  familiar  form  dy/dx  =  Q,  and  the  general  solution  is 

y  =  c  +  j*Qdx. 
If  Q  is  zero,  the  equation  becomes 


and  the  general  solution  is  y  =  ce~JPdx. 
When  Q,  in  the  equation 


is  zero,  the  equation  is  called  the  ordinary  linear  differential 
equation  of  the  first  order  with  the  right  hand  member  zero. 

EXAMPLE.     Find  the  general  solution  of  the  equation 
dy      1 

--     -        * 


Multiply  both  sides  of  the  equation  by 

...^e/^  +  VK 
dx  T  x  y 


24     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

fl-dx 

Let  u  =  2/e^  »    . 


.   .  -v-  =  .     .•  .  u  =     xe*x  +  c. 


Now  &J  *  dz  =  elog  z  =  a;. 


.  •  .  u  =  I  x*dx  +  c  =  -7-4-  c.     .•.v  = 
«^  4n 

It  is  usual  to  solve  an  ordinary  linear  differential  equation  of 
the  first  order  by  substituting  directly  in  formula  (1).  Thus, 
in  the  above  example,  formula  (1)  becomes 


y  =  e-fldxfx*efi;dxdx 


c 
*' 


22.  EQUATIONS  REDUCIBLE  TO  THE  LINEAR  FORM 

A  form  easily  reducible  to  the  linear  form  is 

~ 


where  P  and  Q  are  functions  of  x  and  do  not  contain  y  or  deriv- 
atives. 

Divide  by  yn. 

(>)   .-.y- 
Let  y~n+l  =  u. 


which  is  linear  and  can  therefore  be  solved  by  the  methods  of 
Art.  21. 


EQUATIONS  OF  FIRST  ORDER  AND  FIRST  DEGREE    25 

EXAMPLE.     Find  the  general  solution  of  the  equation 
dy       1  2 

Divide  by  f. 


Let  y~2  =  u. 

»     „ dy      du 

.  • .  —  2y    -7-  =  -T-. 

a#       aa; 

—      --u  2s2 

CtiC         iC  / 

.  • .  u  =  -  2ef^dxfx2e'fldxdx  +  cefldx=  -  2a;3  +  cz2. 


Therefore    —  2^2  +  cx*y*  =  1    is   the  general   solution  of  the 
equation. 

23.    VARIABLES  SEPARABLE 

Sometimes  the  equation  Mdx  -f-  Ndy  =  0  can  be  brought  to 
the  form  Xdx  -\-  Ydy  =  0  where  X  is  a  function  of  x  alone  and 
F  is  a  function  of  y  alone.  In  such  a  case  the  general  solution 
is  evidently 


c  being  an  arbitrary  constant. 

EXAMPLE.     Find  the  general  solution  of  the  equation 

„'  =  <>• 

Divide  by 


Therefore  Vl  —  ^  +  Vl  —  y*  =  c  is  the  general  solution  of  the 
equation. 
3 


26     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

The  process  of  reducing  the  equation  Mdx  -f  Ndy  =  0  to  the 
form  Xdx  -f  Ydy  =  0  is  called  separation  of  the  variables. 

24.  EXACT  DIFFERENTIAL  EQUATIONS  OF  THE  FIRST  ORDER 
AND  FIRST  DEGREE 

Definition.  The  ordinary  differential  equation  Mdx  -|-  Ndy  =  Q 
where  M  and  N  are  functions  of  x  and  y,  is  said  to  be  exact  when 
tthere  is  a  function  u(x,  y)  such  that  du  =  Mdx  -f  Ndy. 

EXAMPLE.  The  equation  2xydx  +  z?dy  =  0  is  said  to  be 
exact  because  u  =  x2y  is  such  that  du  =  2xydx  -|-  z?dy. 

When  there  is  a  function  u(x,  y)  such  that  du  =  Mdx  -f  Ndy, 
then  u  =  c,  where  c  is  an  arbitrary  constant,  is  the  general  solu- 
tion of  the  equation  Mdx  -f  Ndy  =  0. 

Condition  that  the  equation  Mdx  -f  Ndy  =  0  be  exact.  If 
the  equation  Mdx  -{-  Ndy  =  0  be  exact,  then,  by  definition,  there 
is  a  function  u(x,  y)  such  that  du  =  Mdx  -f  Ndy.  Now 

,        du  ,        du  7 
rfw  =  ^- -dx+  —  cfy, 
d#  cty 

from  the  definition  of  the  differential  of  two  independent  variables. 

nr        du  j        TIT        &u 

.'.Jif=^-,     and     &=—. 

dx  dy 

dM       d*u  ,     dN       d*u 

and 


3y       dydx*  dx       dxdy 

dM     dN 
dy   "  dx' 

That  the  equation  Mdx  -f  Ndy  =  0  be  exact,  it  is  therefore 

necessary  that 

dM      dN 

dy  "  dx' 
Conversely,  the  condition  is  sufficient.     That  is  if 

dM      dN 
dy  ==  dx' 

then  Mdx  +  Ndy  =  0  is  an  exact  differential  equation. 


EQUATIONS  OF  FIRST  ORDER  AND  FIRST  DEGREE    27 

Proof:  Let         f Mdx  =  P,     .-.~  =  M. 
J  dx 


'  dy  dx  ~~  dy   "  dx  * 


dx      dy  dx      dx\dy/' 


where  §(?/)  is  such  that  dQ(y~)  =  F(y)dy. 
Therefore,  if 


___ 

^rc  ~~  dy  ' 

the  left  hand  member  of  the  equation  Mdx  +  Ndy  =  6  is  an 
exact  differential  and  therefore  the  equation  is  an  exact  differential 
equation. 

To  find  the  general  solution  of  the  equation  Mdx  +  Ndy  =  0 
when  the  equation  is  exact. 

Let  u(x,  y)  be  a  function  whose  differential  is  Mdx  -{-  Ndy. 

Since  -=-  =  My 

ox 


J>o-     3w     ,T 

/f  Since    -  =  JV; 


Tygxjjtew 

The  general  solution  of  the  equation  is  u  =  c  where  c  is  anxo  v 
arbitrary  constant. 


28     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 


EXAMPLE.     Find  the  general  solution  of  the  equation 
(x*  -f  2xy  +  y)dx  +  (y*  +  x*  +  x^dy  =  0. 

This  is  an  exact  differential  equation.     Therefore  the  general 
solution  can  be  obtained  by  the  above  method. 

Since  —  =  Jf, 

#4 

2xy  +  y*)dx  -f  -F(y)  =  -r  +  x*y  -{-  xy  + 
4 

Now  ^=  JV. 

y  "X 

" '          -  if  4-  x*  4-  «. 


•  '  •  u  =  4  +  xy  +  *y  +  f 

x*  y* 

Therefore  j-  +  x*y  +  xy  +  ~  =  c  is  the  required  general  solution 

of  the  equation.  ^ 

v4^       25.   INTEGRATING  FACTORS 

It  sometimes  happens  that  the  differential  equation 

Mdx  +  Ndy  =  0 

is  not  exact  but  becomes  so  when  it  is  multiplied  by  some  quan- 
tity.    Thus, 


of  Art.  21,  is  not  exact  but  becomes  so  after  multiplication  by 
efpdx. 


EQUATIONS  OF  FIRST  ORDER  AND  FIRST  DEGREE    29 

Definition.  A  factor  which  changes  a  differential  equation 
into  an  exact  differential  equation  is  called  an  integrating  factor 
of  the  equation. 

Sometimes  an  integrating  factor  can  be  found  by  inspection. 

EXAMPLE.     Find  the  general  solution  of  the  equation 
(tfe*  —  y^dx  +  2xydy  =  0. 

The  equation  is  not  exact  as  it  stands  but  becomes  so  on  multi- 
plication by  1/x2. 
Multiply  by  1/x*. 


.  •  .  #  +  -  =  c.     .  •  .  y*  =  —  a*f  ,+  ex. 
x 

Therefore  y*  =  —  xe?  -f  ex  is  the  general  solution  of  the  equa- 
tion. 

Rules  have  been  devised  for  finding  integrating  factors  in 
many  cases  where  they  cannot  be  found  by  inspection.  For  a 
discussion  of  them,  the  student  is  referred  to  Boole's,  Murray's, 
or  Johnson's  Differential  Equations. 

26.    EQUATIONS  HOMOGENEOUS  IN  X  AND  y 

Definition.  If  M  and  N  of  the  equation  Mdx  +  Ndy  =  0 
are  both  of  the  same  degree  in  x  and  y  and  are  homogeneous,  the 
equation  is  said  to  be  homogeneous. 

To  find  the  general  solution  of  the  equation  Mdx  +  Ndy  =  0 
when  the  equation  is  homogeneous. 

dy_       M_ 

dx~  ".AT 


30     SHORT  COURSE  ON  DIFFERENTIAL   EQUATIONS 

Divide  both  numerator  and  denominator  of  —  -~  by  x  raised  to 

the  power  indicated  by  the  degree  of  M  or  N. 

Then  every  term  in  M  and  N  is  constant  or  in  the  forr^  of  a 

coefficient  multiplied  by  some  power  of  - . 
Then 


Let  y  =  vx. 

dv 

•'•xdx  +  v= 
Therefore 

dx  dv 


x      f(v)  —  v9 

an  equation  in  which  the  variables  are  separated,  and  can  ther 
fore  usually  be  integrated  without  difficulty. 

EXAMPLE.       Find    the    general    solution    of   the   equation 
(x2  +  yz)dx  —  xydy  =  0. 


Let  y  =  vx. 

dv      1  4-  v9 

.  •  .  v  4-  x  -y-  =  —  —  .     .  •  .  x  -y-  =  —  '  —  —  v  =  -. 
ax          v  dx          v  v 

,        dx 

.  '  .  vdv  =.  —  . 
x 

.  •  .  v*  =  2  log  ex. 

Therefore  y*  =  2x2  log  ex  is  the  general  solution  of  the  equation. 

dv      a.x  4-  b,y  4-  c. 

27.    EQUATIONS  OP  THE  FORM  ^  =     l     ^  _iy  ^—l 

dx      a^x  -f  o2y  -f  c, 

The  general  solution  of  an  equation  in  the  above  form  can  be 
found  as  follows  : 

Let  x  =  x'  -j-  XQ,  and  y  =  yf  -f  y0,  where  x'  and  y'  are  new 
variables,  and  XQ  and  yQ  are  constants. 


EQUATIONS  OF  FIRST  ORDER  AND  FIRST  DEGREE    31 
Change  the  variables  to  xf  and  y'. 


Case  I.     If  XQ  and  i/0  can  be  determined  such  that 

<¥*<,  +  %o  +  ci  =  °>     and     Vo  +  %o  +  c,  =  0 
then,  on  determining  them  such,  equation  (1)  becomes 


which  is  homogeneous  and  can  be  solved  by  the  method  of 
Art.  26. 

Case  II.     If  x0  and  y0  cannot  be  determined  such  that 
ai*o  +  %o  +  «i  =  0,     and     atxQ  +  b^  +  ct  =  0, 
then,  as  was  seen  in  algebra, 


By  substitution,  the  original  equation  becomes 
%  _       ^  +  %  +  CE 


rfa;       m(a^  +  %)  +  c 
Let  a,#  -|-  6^  =  v. 


Therefore 


dv  ,    v  -f- 

"'    4"*"    1 


an  equation  in  which  the  variables  are  separable. 

EXAMPLE  1.     Find  the  general  solution  of  the  equation 


-  2y  —  7 
;  +  3    -  6' 


32      SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

Let  x  =  x0  +  x',  y  =  yQ  +  y'  where 

6s0_2y0_7  =  0,     and     2*0  +  8y0  -  6  =  0. 


6-2i;  6  -  4i>  - 

""  ~ 


(2  +  3v)dv       <M 
'  "  '  6  —  4v  —  3v2  =  "^r* 

.  •  .  —  log  cta/  =  log  (3v2  +  4v  —  6)*. 


Therefore  3^/2  +  4^  —  60^  —  12y  +  14a;  =  c  is  the  general  solu- 
tion of  the  equation. 

EXAMPLE  2.     Find  the  general   solution   of   the   equation 


This  comes  under  case  II. 

2y  =  *.  g  _  2%  =  *, 

efo       eta* 

•      •    \J   —    ju 


~          ~A    =   ~i~» 

-f  4      oo; 

.  rf«       2v  +  76 
"da;  "'    v  +  8 

•    (^  + 
~ 


.  •  .  v  -  30  log  0  +  38)  =  2x  +  c,. 
.•  .  4*  -  2y  -30  log  (fa  -  2y  +  38)  =  c,. 


EQUATIONS  OF  FIRST  OEDEE  AND  FIEST  DEGEEE    33 

Therefore  2x  —  y  —  15  log  (3x  —  y  -\-  19)  =  c  is  the  general  so- 
lution of  the  equation. 

EXERCISES 

Find  the  general  solution  of  each  of  the  thirty-six  following 
equations. 


>    3.        =a?-y.  4.   (1  +  a:2) 

dy  x  1 

+  =  ' 


6. 


7.   —  +  cos  »  y  =  \  sin  2a;. 


r     9.   -^  +  sin  x  y  =  y2  sin  x.       ^    10.   (1  —  #2)  —  —  xy  =  a.Ti/2. 


11.        +  cos  *</  =  </»  sin  2*.      ^12.  3y>lc 
13.   ~  —  tanxy  =  y*secx. 


14.  y  $?~^ldx  +  a  V2/2  —  Idy  =  0.  _ 

15.  (e8'  -f  1  )  cos  x  dx  +  &  sin  xdy  =  Q. 


16.  V2a?/  —  y*  cosec  x  dx  +  y  tan  x  dy  =  0. 

17.  t/(3  +  ?/)  -j^  =  #(2i/  +  3).  ^. 

rc^  <*& 


18.   (a?  -  Safy  4-  Say1  -  7/)<fo      - 

=  0. 


34     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 


19.  Os  +  <%  +  f)dx  +  (2a?  +  2xy  +  4y*)dy  =  0. 

20.  sin  a:  cos  y  dx  +  cos  #  sin  ydy  =  Q. 


21.  («« 

22.  z(>  -  2y)dy  +  (a;2  +  2i/2)cfo  =  0. 

23.  Saiydy  -  (V  +  ^efe  =  0. 

24.  (z2  +  3^  -  y^dy  -  Stfdx  =  0. 

^  25.  (^  +  2^)dy  -  (3^2  -  2^  +  ifidx  =  0. 

26.  Sicydy  -  (4a;2  +  y«)(fe  =  0. 

27.  (V  —  2a^)dy  +  (V  —  3«y  +  2^)  da?  =  0. 

28.  3ar«dy  +  (2^  _  3y2)da;  =  0. 


29.   (3x  +  2y  -  7)  -^  =  2x  -  By  +  6. 

| 

31.   (5z  —  2y  +  7)  -^  =  a;  —  3y  +  2. 

X*  *  ^ "        rf*  ~  ^ 

^      '     *        ..+      <fa-    ^    ^+    ' 
34.   (5a;  —  2y  +  7)  -j-  =  Wx  —  4y  +  6. 

36.   (6a?  -  4y  +  1)  ^  =  3a;  _  2y  +  1. 

The  following  formulas,  derived  in  almost  any  work  in  cal- 
culus, are  inserted  here  for  convenience  of  reference  : 


EQUATIONS  OF  FIRST  OEDEE  AND  FIRST  DEGREE    35 

The  subtangent  and  subnormal  at  a  point  (#,  ?/)  on  a  curve 
whose    equation   is  expressed    in   rectangular    coordinates    are 

y  -y-  and  y  -~  respectively.     The  polar  subtangent  and  polar  sub- 
normal at  a  point  (r,  0)  on  a  curve  are  r2  -7-  and  -^  respectively. 

The  angle  between  the  radius  vector  to  a  point  (r,  0)  and  the 
tangent  line  to  the  curve  at  the  point  is 


3r" 

dO 

The  equation  of  the  tangent  line  to  the  curve  y  =/(#)  at  the 
point  (xv  2/x)  on  the  curve  is 

dy       ,        N 

|i    y      _    _«y.  //£    /£     ) 

«#   ar=«i 

The  area  enclosed  between  the  curve  y  =/(#),  the  #-axis,  and 
the  ordinates  whose  abscissas  are  XQ  and  xl  respectively  is 


r< 


provided  the  curve  does  not  cut  the  #-axis  between  x0  and  xr 
The  length  of  the  arc  of  the  curve  y  =  /(#)  between  the  points 
(#0,  2/0)  and  (xv  y^  on  the  curve  is 


dx 

37.  Determine  the  curve  whose  subtangent  at  a  point  on  it  is 
fi-times  the  abscissa  of  the  point.     Find  the  particular  curve  that 
goes  through  the  point  (3,  4).     Plot  the  curve  (a),  for  n  =  1, 
(6),  for  n  =  2. 

38.  Determine  the  curve  whose  subtangent  at  a  point  on  it 
is  n-times  the  subnormal  at  the  point.     Find  the  particular  curve 
that  goes  through  the  point  (  ^n,  2).      Plot  the  curve  when 


36     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

39.  Determine  the  curve  whose  subtangent  is  constant  and 
equal  to  a.     Plot  the  curve,  (a),  when  a  =  1,  (6),  when  a  =  2. 

40.  Determine  the  curve  whose  subnormal  is  constant  and 
equal  to  a.     Find  the  particular  curve  that  goes  through  the 
point  (1,  2). 

41.  Determine  the  curve  which  is  such  that  the  length  of  the 
perpendicular  from  the  foot  of  the  ordinate  of  any  point  on  the 
curve  to  the  tangent  line  at  that  point  is  constant  and  equal  to  a. 
Determine  the  particular  curve  when  c  =  a.      At  what  angle 
does  this  curve  cut  the  i/-axis  ? 

42.  Determine  the  curve  which  is  such  that  the  area  between 
the  curve,  the  #-axis,   and  two  ordinates,  is  equal  to  the  arc 
between  the  ordinates. 

43.  Determine  the  curve  which  is  such  that  the  perpendicular 
from  the  origin  upon  any  tangent  line  is  equal  to  the  abscissa  of 
the  point  of  contact. 

44.  Determine  the  curve  in  which  the  angle  between  the  radius 
vector  and  the  tangent  line  is  n-times  the  vectorial  angle.     Plot 
the  curve  when  n  =  \. 

45.  Determine  the  curve  in  which  the  polar  subnormal  is  pro- 
portional to  the  sine  of  the  vectorial  angle. 

46.  Determine  the  curve  in  which  the  polar  subtangent  is  pro- 
portional to  the  length  of  the  radius  vector. 

The  equation  for  a  circuit  containing  induction  and  resistance  is 

Tdi       _. 


where  e  is  the  electromotive  force  [E.M.F.]  impressed  upon  the 
circuit,  R  the  resistance  offered  by  the  circuit,  L  the  coefficient 
of  induction,  i  the  current,  and  t  the  time  during  which  the  cir- 
cuit is  in  operation.  In  each  of  the  four  following  exercises, 
determine  the  current  in  the  circuit  after  a  time  t  supposing  that 
the  resistance  and  induction  are  constant. 

47.  The  E.M.F.  is  zero.     Solve  subject  to  the  condition  that 
i  =  /when  t  =  0. 

48.  The  E.M.F.  is  constant  and  equal  to  E. 


EQUATIONS  OF  FIRST  ORDER  AND  FIRST  DEGREE    37 

49.  The   E.M.F.    is   a   simple   sine   function   of  the   time, 
=  E  sin  at  where  E  is  the  maximum  value  of  the  impressed 
E.M.F.,  and  <o  is  the  angular  velocity,  equivalent  to  2?m  where 
n  denotes  the  number  of  complete  periods  or  alternations  per 
second. 

50.  The  E.M.F.  is  the  sum  of  two  components  each  follow- 
ing the  sine  law,  that  is,  e  =  E^  sin  ut  -\-  Ez  sin  (but  -f-  0). 

The  equation  for  a  circuit  containing  resistance  and  capacity  is 

di         i         1  de 


where  e  is  the  E.M.F.,  H  the  resistance,  C  the  capacity,  i  the 
current,  and  t  the  time  during  which  the  circuit  is  in  operation. 
In  each  of  the  two  following  exercises  determine  the  current 
in  the  circuit  after  a  time  t,  supposing  that  the  resistance  and 
capacity  are  constant. 
¥     51.  The  E.M.F.  is  constant  and  equal  to  E. 

52.  The   E.M.F.    is  a   simple   sine  function   of  the  time, 
=  E  sin  wt. 

The  equation  for  a  circuit  containing  resistance  and  capacity  is 

Rd(l  ,    V       e 
*dt+~C  = 

where  e  is  the  E.M.F.,  R  the  resistance,  C  the  capacity,  q  the 
quantity  of  charge  in  the  conductor,  and  t  the  time  during  which 
the  circuit  is  in  operation.  In  each  of  the  three  following  exer- 
cises determine  the  charge  in  the  circuit  after  a  time  t,  suppos- 
ing that  the  resistance  and  capacity  are  constant. 

53.  The  E.M.F.  is  zero.     Solve  subject  to  the  condition  that 
q  =  Q  when  t  =  0. 

54.  The  E.M.F.  is  constant  and  equal  to  E. 

55.  The   E.M.F.    is   a   simple   sine  function   of  the  time, 


ANSWERS 

sc 
1.  4xy  =  2#2  —  x*  +  c.  2.  y  sin  x  =  log  tan  -  +  c. 


SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

y  =  x  —  1  +  ce~*.  4.  y  =  x  —  1—  tan~1z+cetan~lx 


y=  —  ^  log  (1  —  a2)  +c#.     7.  y  =  sin  x  —  1  + 
« 

_  i 

y  =  az  +  ca;  VI  —  x*.  9.   -  =  1  +  ce"00335. 

y 


=  2  sin  a  - 


-(1-w)  8in  x 


38 

3. 
5. 

6. 

8. 
10. 

11. 

12. 
13. 

14. 
15. 

16. 
17. 
18. 

19.    '^-  +  2x*y  +  ^2  +  7/4  =  c.      20.  cos  x  cos  y  =  c. 
x* 


t/3  =  x  —  2  +  ce~*. 

y~*  =  —  3  sin  #  cos2  .^  —  sin3  a?  -j-  c  cos3  x. 

V^2  —  1  —  sec"1  a;  +  A&^—  1  —  sec"1!/  =  c. 
(^  _^_  1)  sin#  =  c. 

coseco;  +  V2ay  —  y2  —  avers"1-  =  c. 

a 

2^2  +  6y  -  9  log  (2y  +  3)  =  4z2  +  e. 


21.    —  +  x  log  2/  =  c. 

d 

23.  (4/  -  >•)•  -  «•- 


22.  a;2ex  =  c(x  +  t/)8. 
24.  y  .  c 


26.   (a;2  -  ^)5  = 


(3 


27.  y  =  *log-. 

X 


EQUATIONS  OF  FIRS1   ORDER  AND  FIRST  DEGREE 

28.   ex  =  |6y-(3  +  V33>1^ 


29.  (y  -  f|)2  +  3(s  -  A)  (y  -  f  f  )  -  (*  _  A)2  =  c. 

30.  (5y  -  2x  -  3)4  =  c(4y  -  4*  _  3). 

(2(y  -  A)  -  (4  - 

31 


-  T83)  -  (4 

32.  (3y  -  5«  +  10)2  =  c(y  -  a?  +  1). 

33.  15y  -  30o;  +  c  =  3  log  (5x  -  15y  +  17). 

34.  4#  _  2y  +  c  =  16  log  (5a?  —  2y  +  23). 

35.  2y  —  a?  +  c  =  log  (x  —  y  -f  2). 

36.  2y  -  x  +  c  =  J  log  (12«  -  8y  +  1). 

37.  if  =  ex  ;  yn  =  **x.  38.  y=---  a?+c;  y=  — 

\^  '  W 

39.  7/a  =  ce1. 

40.  y2  =  2ax  -f  c  ;  y2  =  2ao;  +  4  —  2a. 

C/—        tt2_^.\  tt       ^  __  * 

41.  y  -  ^  f  6-  +  ^  e   «    ;  y  =  -  (e«  +  e~^)  ;  zero. 

*  \  I.  * 

42.  y  =  ^'(^*  +  cVfca:).  43.  a;2  4.  f  =  ex. 
44.  rn  —  c  sin  n0.                             45.   r  =  c  —  &  cos  0. 
46.  r  =  cee/k.                                  47.  i  =  Ie~^e. 

48.  t  =  —  +  ce~z*. 

49.  t  =  —  j-^  --  r  (  -=-  sin  w^  —  o)  cos  out]  +  ce~~Lt  . 


40     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 
50.  i  =  —  7-™—  -  r-  (  >  sin  o>£  —  w  cos  <»t  } 


r  /?  ~i 

) 


+  ce   L  . 

51.  i  =  ce~^t . 

-  — t  CE& 

52.  i  =  ce  RG  4-  ^ ,ir™-«  (cos  w<  -h  -KCw  sin  o>0. 

r  1  +^2CVV 

53.  5=  Qe'*0'.  54.   g=  CE+cJT*** 
55.  9  =  1— 


CHAPTER  IV 

ORDINARY  LINEAR  DIFFERENTIAL  EQUATIONS  WITH 
CONSTANT   COEFFICIENTS 

28.  Definition.  An  ordinary  linear  differential  equation  is 
a  differential  equation  in  one  dependent  variable  which  is  linear 
in  the  dependent  variable  and  its  derivatives. 

We  saw  in  Art.  21  that  the  type  of  an  ordinary  linear  differ- 
ential equation  of  the  first  order  is 


where  P  and  Q  are  functions  of  x,  and  do  not  contain  y  or  de- 
rivatives. 

In  general,  the  type  of  an  ordinary  linear  differential  equa- 
tion is 

dny       _  dn~ly       _  dn~*y 

' 


where  Pv  P2,  •  •  •  ,  Pn,   and  X,  are  functions  of  x,  and  do  not 

contain  y  or  derivatives. 

In  this  chapter  the  only  cases  considered  are  those  where 
Pj,  P2,  •  •  •  ,  Pn  are  constants  and  real.  Two  forms  of  this  equa- 
tion present  themselves,  namely,  when  the  right  hand  member  is 
zero,  and  when  the  right  hand  member  is  not  zero. 

RIGHT  HAND  MEMBER  ZERO 

29.  We  shall  first  prove  a  theorem  used  in  the  investigation 
of  equations  in  this  form.  It  is  : 

Theorem.  If  y  ==  yv  y  =  y2,  •  •  •  ,  y  =  yn,  are  solutions  of  the 
equation 


41 


42     SHOET  COURSE  ON  DIFFERENTIAL  EQUATIONS 
then 


arbitrary  constants,  is  also  a  solution  of  the  equation. 

Proof.     Substitute  y  =  clyl  -f-  c^,  -f  •  -  -  +  c^  in  the  equa- 
tion. 


Now  each  expression  in  brackets  is  zero,  since 

y  =  Vv  y  =  y,>  •  •  •>  y  =  </»> 

are  solutions  of  the  equation.     Therefore 


is  a  solution  of  the  equation. 

Cor.  If  y  =  c^  _f-  c2i/2  +  -  •  •  +  cnyn  is  a  solution  of  the  equa- 
tion, then  y  =  c^yv  y  =  c,i/2,  •  •  -  ,  y  =  cn^n,  are  solutions  of  the 
equation. 

If  y  =  yv  y  =  t/2,  •  •  •,  y  =  yn  are  linearly  independent  solu- 
tions of  the  equation,  then  y  =  c^  -f  c3y2  -f  •  -  •  +  cnyn  is  the 
general  solution  of  the  equation  (see  Art.  13). 

30.  To  find  a  solution  of  the  equation 


in  the  form  ^/  =  ^nx- 

Let  y  =  e™  and  substitute  in  the  equation.     If  y  =  e™*  is  a 
solution  of  the  equation,  then 

-f  PX1-1  +  Pamn~2  +  •  •  •  +  PJ  =  0. 


LINEAR  EQUATIONS  WITH  CONSTANT  COEFFICIENTS  43 
Since  emx  cannot  be  zero  for  any  value  of  m,  then  must 
mn  +  P1"1  +  P2mn~2  +  •  •  •  +  P  =  0. 


Therefore,  if  y  =  emx  is  a  solution  of  the  equation,  it  is  necessary 
that 

mn  +  PX1'1  +  P2mn~2  +  •  •  •  +  Pn  =  0. 

Conversely,  if  m  has  a  value  m^  such  that 

mf  +  P,m^  +  P2?V~2  +  •  •  '  +  -P,  =  0, 

then  y  —  emix  is  a  solution  of  the  equation.     This  is  obvious  be- 
cause on  substitution  of  y  =  emix,  the  equation  reduces  to 

0*i*  (m»  +  P^"-1  +  Pjtn,?~*  +  •  •  -  +  PJ  =  0. 

Therefore  the  necessary  and  sufficient  condition  that  equation  (1) 
has  a  solution  in  the  form  y  =  e™*  is  that  m  be  such  that 


mn  +  PX"1  +  -PjWi1*"1  +  •  •  •  +  P*  =  0. 
Definition.     The  equation 

mw  +  PfriT*  +  P2mn~J  +  •  •  -  +  Pn  =  0 
is  called  the  auxiliary  equation  of 


31.   To  find  the  general  solution  of  the  equation 

n^^ 

= 


When  the  auxiliary  equation  has  distinct  roots.  Denote  the 
roots  by  mv  m2,  •  •  •,  mn.  Then  n  linearly  independent  solu- 
tions of  the  equation  are  y  =  e™1*,  y  =  e™2*,  •  -  •,  y  =  emnX,  and 
the  general  solution  is  y  =  c^*  +  c^x  -f  •  •  •  +  cnem^  (see 
Art  29). 


44     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 
EXAMPLE  1.     Find  the  general  solution  of  the  equation 


_--. 

dx*         dx 

Let  y  =  eP*. 

.  •  .  e-'Cm8  -  3ro  _  4)  =0.     .*.  (w_4)(w+  1)  =0. 

Therefore  y  =  c^  -j-  c^e"*  is  the  general  solution  of  the  equation. 
EXAMPLE  2.     Find  the  general  solution  of  the  equation 


. 

dx*   '   dx   ' 

Let  y  =  (?**. 

+  m  +  4)  =0. 


1-V15A/  1   +  VT5 

--  ~  ~~ 


^ 
Therefore  y  =  c^       a     *  -f  cae       a     *  is  the  general  solution  of 

the  equation. 

This  solution  may  be  written  as 

y  =  cf**  cos  —  ^—  #  +  cse~*x  sin  —  ^-  x     (see  Art.  5). 

When  the  auxiliary  equation  has  multiple  roots.  Suppose 
that  the  auxiliary  equation  mn+P1mw~1-f  P2mn~2  -f-  •  -  •  -j-Pn  =  0 
has  the  roots  mv  m2,  w3,  •  •  •  ,  mn  . 

At  first  suppose  that  two  roots  are  equal.  Suppose  for  defi- 
niteness  that  m2  =  m,.  Then  a  solution  of  the  differential  equa- 
tion is 

y  =  (Cj  +  c2)emi*  +  C3em3*  -f  c^*. 

Since  c,  +  c2  is  equivalent  to  only  one  constant,  this  solution  con- 
tains only  Ti  —  1  arbitrary  constants  and  is  not  therefore  the  gen- 
eral solution  of  the  equation. 

To  find  the  general  solution  in  this  case  : 


LINEAR  EQUATIONS  WITH  CONSTANT  COEFFICIENTS  45 

Suppose  that  the  differential  equation  is  such  that  its  auxiliary 
equation  has  the  roots  mv  m^  -f  A,  w3,  •  •  • ,  ran.  The  general 
solution  of  this  equation  is 


y  = 


Expand  d™  by  Maclaurin's  Theorem  to  n  terms  and  the  re- 

mainder. 

1  *-1       (A«)w 

+"  ' 


Substitute  in  the  above  equation. 


0  <  xl 


-l.l 

6   J 


Since  c2  is  arbitrary,  A  may  be  chosen  such  that  cth  is  any  con- 
stant B  for  all  values  of  h.  Since  c,  is  arbitrary,  cx  -}-  c2  may  be 
chosen  such  that  ct  +  c,  =  -4.  Then 


(1) 

where  A  and  -B  are  arbitrary  constants. 

Let  h  approach  zero.  As  h  approaches  zero,  the  assumed 
auxiliary  and  differential  equations  approach  identity  with  the 
given  ones,  and  (1)  approaches  the  general  solution  of  the  given 
differential  equation. 

Now 

limit     ** 


+" 


Therefore  the  general  solution  of  the  differential  equation  is 
y  =  (  A  -f-  Bx)em*x  -f  csem3a:  +•••-)- 


46     SHORT  COURSE  ON  DIFFERENTIAL   EQUATIONS 

or,  as  we  shall  write  it, 

y  =  (ct  -f  c^x)emix  -f  cseW3*  _^_  .  .  .  _j_  cnem«x. 

In  a  similar  manner  it  can  be  shown  that  if  three  roots  of  the 
auxiliary  equation  are  equal,  the  general  solution  of  the  differen- 
tial equation  is 


and,  in  general,  if  r  roots  are  equal,  the  general  solution  of  the 
equation  is  % 


If  a  pair  of  imaginary  roots  occur  twice,  the  part  of  the  general 
solution  derived  from  these  roots  is 


y  =  (Cj  +  c2zXa+^>  +  (cs 

sfix+j  sin  /8^)  -f  (^+C/?)«^*(608  /3x—  j  sin  /?x 
?)  cos  jftc  +  (Az  -f  J5,a?)  sin  px]  . 
EXAMPLE  1.     Find  the  general  solution  of  the  equation 


The  auxiliary  equation  is  m2  +  2m  +  1  =  0,  or  (m  -f  I)2  =  0. 
The  general  solution  is  therefore  y  =  e"*^  +  cjc). 

EXAMPLE  2.     Find  the  general  solution  of  the  equation 


The  auxiliary  equation  is  m*  —  4ms  +  8m*  —  Sm  +  4  =  0,  or 
{(m  _  1  _,/)<>  _  1  +./)}'  =  0. 
The  general  solution  is  therefore 

y  =  ee{(Al  +  S^)  cos  a?  +  (A9  +  J52a;)  sin  x}. 


32.  As  a  physical  application  of  the  above  principles,  con- 
sider the  following  discussion  (see  Emptage,  Electricity  and 
Magnetism,  page  180)  : 


LINEAR  EQUATIONS  WITH  CONSTANT  COEFFICIENTS    47 

In  a  galvanometer  in  which  resistance  is  offered  to  the  motion 
of  the  needle,  the  equation  of  motion  of  the  needle  for  small 
oscillations  may  be  written  as 

^  +  2*5+A«—  )-0,  •     (1) 

where  0  is  the  angle  through  which  the  needle  turns  in  the  time 
t,  k  is  a  constant  depending  on  the  resistance  offered  to  the  mo- 
tion of  the  needle,  o>2  is  a  constant  depending  on  the  moments  of 
the  restoring  forces  on  the  needle,  and  a  w  the  angle  which  the 
needle  at  rest  makes  with  the  line  from  which  angles  are  meas- 
ured. Let  0  —  a  =  6',  and  substitute  in  (1). 


This  is  a  linear  differential  equation  of  the  second  order  with 
constant  coefficients  and  right  hand  member  zero.  The  auxiliary 
equation  is  m2  -f  2km  -f-  o>2  =  0.  The  roots  of  the  auxiliary 
equation  are  m  =  —  k  ±  V&2  —  <*>2- 

Case  I.     Jfk>  o>. 

In  this  case,  0  -  a  =  c1rf-*+v  '**-«**  +  c8eC-*-y*2-«2)<  is  the 
general  solution  of  (1). 

Case  II.     If  k  =  w. 

In  this  case  0  —  a  =  (cx  +  c2<)e~w  is  the  general  solution  of 
(1). 

Case  III.     If  k  <  o>. 

In  this  case  0  —  a  =  e"**^  cos  V«»2  —  kz  t  +  c2  sin  V«>2  —  &2  (] 
is  the  general  solution  of  (1). 

In  cases  I  and  II  the  motion  is  not  oscillatory.  The  needle 
can  go  through  the  position  of  equilibrium  for  one  value  of  t, 
after  which  it  reaches  a  position  of  maximum  deflection  and  then 
continually  approaches  but  never  reaches  the  position  of  equi- 
librium. In  case  III  there  are  oscillations  in  equal  times,  the 
periodic  time  being 


48     SHORT  COURSE  ON  DIFFERENTIAL   EQUATIONS 

EXERCISES 
Find  the  general  solution  of  each  of  the  following  equations. 


-*+*-«.    4.  *+--*.« 

^ 

dsv      «^2V       o^V  *      n    d*y      nd*y      dy 

.   j4  —  3  -j-5  +  3  -/  —  y  =  0.     6.  -j  ^  +  3  -^  +  -/•  —  5y  =  0 
rfor1         da?  n      rfa;      y  d^         eta2  T  rfa; 


^/  ANSWERS 

1.  y  =  c^"1*  +  c,6"«.  2.  y  =  clc 

3.  y  =  c^  +  c^x  +  cy  4.  y  =  e^ 

5.  y  =  e'Oj  +  c2a;  +  csa^). 

6.  y  =  ^e*  +  e~2*(ca  cos  a?  +  cs  sin  #). 

7.  y  —  cf^x  +  ez(c2  cos  a;  -|-  c$  sin  x). 

8.  y  =  Cje35  4-  c2e~*  +  cs  cos  a?  -f  c4  sin  «. 

9.  y  =  (Cj  4-  c2o;)  cos  x  -\-  (cs  -j-  c4#)  sin  a;. 

RIGHT-HAND  MEMBER  NOT  ZERO 

33.  Symbolic  form  of  equation.     The  equation,  when  the  right 
hand  member  is  not  zero,  is 

n  ^  "-'  _ 

+  »y        '          W 


where  P,,  P2,  •  •  •  Pn,  are  constants,  and  Xis  a  function  of  x  but 
not  of  y. 
Let 

d 


LINEAR  EQUATIONS  WITH  CONSTANT  COEFFICIENTS  49 

and  the  equation  may  be  written, 

Dny  _|_  PJ)n~ly  +  P2Dn~*y  +  •  •  •  +  Pny  =  X. 

Suppose  that  y  is  treated  as  an  algebraic  factor  of  the  left  hand 
member  of  the  equation.  On  this  supposition,  the  equation 
becomes 

»-...  + P.  )y  =  X. 


Suppose  also  that  D"  +  P^"'1  +  PJ)n~*  +  •  •  •  +  Pn,  factored 
as  an  algebraic  expression  in  D,  is 


and  that  the  equation  is  written 

X.  (2) 


Equation  (2)  is  not  equivalent  to  equation  (1)  except  in  a 
symbolic  sense.  Let  us  see  what  conventions  must  be  made  in 
order  that  equation  (2)  be  equivalent  to  equation  (1). 

Let  us  make  the  convention  that  (D  —  m)u  where  u  is  a  func- 
tion of  x  is  equal  to 

du 

-j-  —  mu. 
dx 

Also,  let  us  agree  that  we  shall  begin  at  the  right  of  the  left 
hand  member  of  (2)  and  work  towards  the  left,  evaluating 
according  to  the  preceding  convention  at  each  step.  Then 

/r,  N        dy 

- 


(D  -  m^XD  -  mjy  =  (D  _  m^ 

d  (dy 

=  5iU~- 

d*y       ,  .dy 


50     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 
and  finally, 


dny  N  dn 


since  the  factors  of  Dn  +  P^1  +  P2Dn-2  +  •  -  -  +  Pn,  treated 
as  an  algebraic  expression  in  D  are  D  —  mv  D  —  raa,  •  •  •,  D  —  mn. 
This  expression  is  therefore  the  same  as  the  left  hand  member  of 
equation  (1).  Therefore,  with  these  conventions,  equations  (2) 
is  equivalent  to  equation  (1). 

EXAMPLE.     With  the  above  conventions,  the  equation 


may  be  written  in  the  equivalent  form  (D—m^^D—m^y  =  Xt 
where  D  —  ml  and  D  —  ra2  are  the  factors  of  the  expression 
D2  +  PJ)  -f  P2  treated  as  an  algebraic  expression  in  D.  For, 

f/7/ 

(D  -  mjy  =  ^-m#, 
(D  -  m,)(D  -  mjy  =  (D  -  m,)         _ 


Now  —  (ml  -}-  m2)  =  P1?  and  w^m,  =  P2.     Therefore  the  second 
form  of  the  equation  is  equivalent  to  the  first. 

Definitions.  When  equations  in  the  form  (1)  are  expressed 
in  the  form  (2),  they  are  said  to  be  expressed  symbolically,  or 
to  be  expressed  by  means  of  symbolic  factors. 

When  a  symbolic  factor  D  —  m  and  a  function  u  are  applied 

y-7|» 

to  each  other  so  as  to  give  (D  —  m)u  or  -y-  —  .  muy  the  function 

u  is  said  to  be  operated  upon  by  D  —  m,  or  the  factor  D  —  m  to 
be  multiplied  symbolically  by  u. 


LINEAR  EQUATIONS  WITH  CONSTANT  COEFFICIENTS    51 

The  factor  D  —  m  is  called  the  symbolic  operator,  or  more 
briefly,  the  operator. 

34.  Theorem.  The  order  in  which  the  symbolic  factors  in  the 
equation  of  the  last  article  are  taken  is  immaterial. 

Consider  in  illustration  the  equation  of  the  second  order. 
Let  the  equation  be  taken  in  the  form 


Then 

(D  —  m^y  =  -/x  —  miy> 
and 

X  %J  \  I/ •/  V  *         \     f^rf  1«     / 

\ax  i 

d*y  N  c?v 


Therefore  (D  —  w3)(D  —  m^)y  =  X  is  equivalent  to 

3+*2+*-* 

Also,  (Z>  —  ml)(Z>  —  m2)i/  =  X  is  equivalent  to 

g+P^  +  P^X         (See  Art.  33.) 

Therefore,  in  the  case  of  the  equation  of  the  second  order,  the 
order  in  which  the  factors  are  taken  is  immaterial. 

The  proof  in  the  general  case  is  left  as  an  exercise  to  the 
student. 

35.  First  method  of  solution  of  the  equation 

(D-m^CD-m^y  =  X. 
Let  (Z>  —  w2)i/  =  u.     The  equation  then  becomes 

(D  —  m^)u  =  X    or     -p  —  mji  =  X 


52     SHORT  COURSE   ON  DIFFERENTIAL  EQUATIONS 

The  general  solution  of  the  equation 

du  ^ 

dx 
is  (see  Art.  21) 

u  =.  emix  j  < 
.  • .  (D  —  wa)y  =  emix  J  t 


=  e^x  J*  #*-» 
This  is  the  general  solution  of  the  given  equation. 
EXAMPLE.     Find  the  general  solution  of  the  equation 


-- 

dx2         dx  T 

Write  the  equation  as  (Z>  —  1)(Z)  —  2)y  =  cos  a;. 
Let  (D  —  2)y  =  u.     The  equation  then  becomes 

(D  —  l)w  =  cos  a;     or    -=-  —  it  =  cos  «. 
aa; 

.  •  .  u  =  e*  j*  e~*  cos  #  dx  +  c^ 
=  J(sin  #  —  cos  a;)  +  c^e*. 
.  *  .  (D  —  2)y  =  ^(sin  a;  —  cos  x)  -|-  c^. 
.  •  .  y  =  JeJz  JV2*(sin  a;  —  cos  #)cfo  +  c^235  J*  e"*^  +  c,eto 

=  y1^  cos  x  —  y3^  sin  a;  -f-  c^  -f-  c2e2*. 
This  is  the  general  solution  of  the  given  equation. 
36.  To  solve  the  equation 

(D  _  m^(D  _  mj  •  •  •  (D  _  mjy  =  X, 

we  may  proceed  as  follows  : 

First,  let  (D  —  w2)  -  •  •  (D  —  mn)y  =  -u.     The  equation  then 
becomes  (Z>  —  m1)w  =  X. 


LINEAR  EQUATIONS  WITH  CONSTANT  COEFFICIENTS  53 

From  this  equation,  u  can  be  determined  as  in  the  case  of  the 
equation  of  the  second  order.     Let 

(D  _m3)  ...  (D_mn)y  =  v. 

Then  ,  -^          N 

(Z)  —  m^)v  =  u. 

From  this  equation  v  can  be  determined  in  the  same  manner  as 
u  was  determined  before.  After  n  —  1  such  steps  there  results 
(Z)  —  mn  )y  =.  z  where  z  is  a  known  function  of  x. 

The  general  solution  of  the  equation  (Z>  —  mn)y  =  z  is  the 
general  solution  of  the  original  equation. 

37.  The  following  theorems  concerning  the  symbolic  operator 
will  now  be  established  : 

Theorem  I.     A  constant  factor  in  a  function  may  be  written 
in  front  of  the  operator. 

Proof:  Let  au  be  a  function  containing  a  constant  a  as  a  fac- 
tor.    Let  D  —  m  be  the  operator.     Then 

(Z)  —  m)au  =  -=—  —  mau,  by  definition 


/du  \ 

=  a  [  -7-  —  mu  ] 

\d*          I 


=  a(D  —  m)u. 

Theorem  II.  The  result  when  the  operator  is  applied  to  the 
sum  of  a  number  of  functions  is  equal  to  the  sum  of  the  results 
found  when  the  operator  is  applied  to  each  of  the  functions 
separately. 

Proof:  Let  u-\-v  +  w-\----+zbQ  the  sum  of  a  number  of 
functions.  Let  D  —  m  be  the  operator.  Then 

(Z>  —  m)  (u  -\-  v  -|-  w  -\-  •  •  •  +  z) 

d(u  a.  v  4-  w  -L  •  •  •  -L  3) 
=  -^—        ^      ^  -^  —  m(u  +  v  +  w  +  ------  (-2) 


du  dv  dw  dz 

=  -=-  —  mu  4-  -y-  —  mv  4-  -=—  —  mw  -4-  •  •  •  4-  -=-  —  mz 
dx  r  dx  r  dx  "  dx 

=  (D—  m)^+  (Z>—  m)v  +  (Z)  —  m)w;  +  •  •  •  4-  (D  —  ra)z. 


54     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

38.  The   equation  (D  —  mj  (D  —  wa)  •  -  -  (D  —  mn)*/  — 
may  be  written  in  the  form 


In  the  first  form  the  symbolic  operators 

D  —  mv  D  —  mv  -  -  •  ,  D  —  mn 

applied  in  succession  give  X.  Moreover,  by  the  theorem  of 
Art.  34,  the  order  in  which  the  operators  are  applied  is  imma- 
terial. If  the  second  form,  therefore,  is  to  be  the  same  as  the 

first,  the  symbolic  expression  7^  -  ^^r  -  r  -  7^  -  r-  X 

--        •  •  •       - 


7^  -  r- 
(D-roJ 

must  be  such  that,  when  operated  upon  by 

D  —  mt,  D  —  m2,  •  -  •  ,  D  —  ran, 
in  succession  in  any  order,  the  result  is  X. 

Definition.     The  symbol  7-^  -  ^7^  -  r  --  7^  -  -  is 
(D  -  mJ(D  _  m2)  .  •  .  (D  _  mn) 


called  the  inverse  symbolic  operator,  or,  more  briefly,  the  in- 
verse operator. 

39.  Let 


be  a  linear  differential  equation  where  the  symbolic  factors  viewed 

as  algebraic  factors  are  distinct.     Break  up  -^=- 

P 


into  partial  fractions  as  if  it  were  an  algebraic  expression  in  D. 
Then 

_  1_  _  ^_J_     /__!_  O-Y 

(D  —  mx)  (Z>  —  ??i2)       ??ix  —  ma  \  D  —  m1      D  —  mt  / 
Let 

A  !  !        V 

=  u     and     —  -  r=r  -  X  =  v. 


Theorem.       The     result     of     operating     on      u  -{-  v     with 
(D  —  m^(D  —  m2)  is  X 


LINEAR  EQUATIONS  WITH  CONSTANT  COEFFICIENTS   55 

Proof:  Operate  on  u  -f  v  with  (Z>  —  mj)(Z>  —  mt). 

ma)(w  +  v)  =  (D  —  m^  (D  —  m,>  + 


(Z>  —  w^D  —  m,>,  by  theorem  II,  Art.  37. 
Now 

(D  -  m.>  =  -  -  -  X,     and     (D  -  m>  =  _  —  1  -  X 
mx  —  m2  m,  —  w2 

by  definition  and  theorem  I,  Art.  37. 


=  (D  -  ma)  —  -  -  X  +  (D  _  mj     - 
^  y 


40.  When  the  symbolic  factors  D  —  ml  and  D  —  m2,  viewed 
as  algebraic  factors,  are  distinct,  the  result  of  operating  on 


v 


with  (D  —  m1)(D  —  ?/i2)  is  JT,  by  the  preceding  article,  and  the 
result  of  operating  on 


with  the  same  factors  is  X,  by  definition.  Therefore  when  the 
symbolic  factors  D  —  ml  and  D  —  m2,  viewed  as  algebraic  fac- 
tors, are  distinct,  the  inverse  operator  of 


may  be  broken  up  into  partial  fractions  the  same  as  if  it  were  an 
algebraic  expression  in  D,  and  the  result  of  operating  with 
(D  —  mj)(Z)  —  m2)  on  the  expression  formed  by  multiplying 
each  of  the  fractions  symbolically  by  JT,  and  taking  the  algebraic 
sum  of  the  results,  is  X. 


56     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

In  general,  when  the  symbolic  factors 

D  —  mv  D  —  mv  -  •  •  ,  D  —  mn, 
viewed  as  algebraic  factors  are  distinct,  the  inverse  operator  of 


can  be  broken  up  into  partial  fractions  the  same  as  if  it  were  an 
algebraic  expression  in  D,  and  the  result  of  operating  with 
(D  —  mt)  (D  —  m,)  -  -  •  (D  —  mn  ),  on  the  expression  formed  by 
multiplying  each  fraction  symbolically  by  X  and  taking  the 
algebraic  sum  of  the  results  is  X. 

The  proof  of  this  theorem  is  left  as  an  exercise  to  the  student. 

41.  Second  method  of  solution  of  the  equation 


Break  up  7-^  -  ^rn^  -  r  into  partial  fractions  the  same  as 
r 


if  it  were  an  algebraic  expression  in  D. 

*  _!__/_!_        1    V 

(D  _  mj  (Z>  —  ra2)       ml  —  ma  \  D  —  mx      J)  —  m,  / 


Let 


u  _ ^^ X    and    v  =  —  • 


Operate  on  u  with  D  —  m,. 

dii  1        TT 

.  * .  -j-  —  m.u  = X. 

ax  ml  —  ma 

i  r 

.  • .  u  =  -       —  e^i*  I  e-77*!*^^ 
mi  —  ma        J 

Operate  on  v  with  D  —  m, 

*>  1  XT 

.•.-——  m,v  =  — X. 

ax  ml  —  ra2 

.  • .  v  =  -  — - — -  <P*   Ce 
m1  —  mt        J 


LINEAR  EQUATIONS  WITH  CONSTANT  COEFFICIENTS    57 

.  •  .  y  =  -  emi*  Ce-™*xXdx  —  -  em*x  Ce~m*xXdx 
ml  —  mt        J  m1  —  ra2        J 

-|_  c1emi:B  -f 
which  is  the  general  solution  of  the  equation. 

EXAMPLE.     Find  the  general  solution  of  the  equation 


Write  the  expression  in  the  form 


Break  up  YT\  -  TvTn  -  o\  ^nto  Par^al  fractions  the  same  as  if 
(U  —  L)\U  —  *) 

it  were  an  algebraic  expression  in  D. 

1  11 


Let 


1  ,1 

— -  cos  x  =  u    and    yr — ^  cos  x  =  v. 

—  1  U  —  £ 


Operate  on  u  with  D  —  1. 

du 

.  • .  _  _  -w  —  _  cos  a;. 
a# 

.  • .  ^  =  ^  cos  x  —  J  sin  x  -f-  ^e*. 
Operate  on  v  with  D  —  2. 

dv       0 

.  • .  -j-  —  2v  =  —  cos  #. 
dx 

.• .  v  =  —  f  cos  x  +  |  sin  a?  +  c2e2*. 
.'.y  =  yV  cos  re  —  y3^  sin  a;  -f-  c^  +  C2e2x, 

which  is  the  general  solution  of  the  equation. 

This  method  does  not  apply  when  the  symbolic  factors  viewed 
as  algebraic  factors  are  not  distinct. 
5 


58     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

42.  It  will  be  noticed  in  the  example  of  the  preceding  article 
that  the  result  is  the  same  as  that  found  by  applying  the  method 
of  Art.  35  to  the  same  equation.     This  will  be  the  case  in  any 
linear  differential  equation  with  constant  coefficients  to  which 
both  methods  apply. 

The  first  method  of  solution  will  apply  in  all  cases  where  the 
left  hand  member  of  the  equation  can  be  factored  into  linear  fac- 
tors in  D.  The  second  method  will  also  apply  if  the  linear  fac- 
tors in  D  are  all  distinct.  If  two  or  more  factors  are  equal,  and 
the  inverse  operator  be  broken  up  into  partial  fractions,  the  term 
or  terms  corresponding  to  these  factors  may  be  evaluated  by  the 
first  method. 

Usually  the  second  method  is  easier  of  application  than  the 
first. 

43.  An  examination  of  either  method  by  which  the  general 
solution  of  a  linear  differential  equation  of  the  nth  order  with 
constant  coefficients  and  second  member  not  zero  is  derived  shows 
immediately  that  the  general  solution  consists  of  the  sum  of  two 
parts,  one  containing  terms  not  involving  arbitrary  constants, 
the  other  containing  terms  involving  such  constants.  •  Moreover 
the  arbitrary  constants  are  involved  so  that  when  any  one  is  zero, 
the  term  in  which  it  appears  vanishes. 

Definition.  The  part  of  the  general  solution  of  a  linear  dif- 
ferential equation  with  constant  coefficients  and  second  member 
not  zero  which  contains  the  arbitrary  constants  is  called  the  com- 
plementary function  of  the  general  solution  of  the  equation. 

EXERCISES 

Find  the  general  solution  of  each  of  the  fourteen  following 
equations. 


4. 


LINEAR  EQUATIONS  WITH  CONSTANT  COEFFICIENTS   59 

dzy         d*y         dy 

fc*r 


7  2 

7      -  J 


0     d2y      ft  dy 
8.    ^  —  2  -/  +  y  =  re. 
2 


4  ^ 

10.    -j+ 


.  dy 
- 


11      y  I  y  _  sjn  x 

dx2^~  y 


12.       - 


—y=cosx. 


'dx' 


In  each  of  the  six  following  exercises,  find  the  equation  of  the 
elastic  curve  of  the  beam  from  the  given  differential  equation, 
determining  the  constants  of  integration.  Find  also  the  deflec- 
tion of  the  beam.  In  these  equations,  E  is  the  modulus  of  elas- 
ticity, I  is  the  moment  of  inertia  of  a  cross  section  of  the  beam 
about  a  gravity  axis  in  the  section  perpendicular  to  the  applied 
forces,  and  /  is  the  length  of  the  beam. 

15.  The  beam  rests  on  supports  at  its  ends.  It  is  supposed 
weightless  with  a  weight  P  at  its  middle  point. 


60 


SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 


16.  The  beam  rests  on  supports  at  its  ends.     It  is  supposed  to 
be  of  uniform  cross  section  and  of  weight  w  per  unit  of  length. 

Yl 


da?'' 


17.  The  beam  rests  on  supports  at  its  ends.  It  is  supposed  to 
be  of  uniform  cross  section  and  of  weight  w  per  unit  of  length, 
and  to  have  a  weight  P  at  its  middle  point. 

Y 


0     X 


18.  The  beam  is  a  cantilever  fixed  horizontally  in  the  wall. 
It  is  supposed  weightless  with  a  weight  P  at  its  extremity. 

Y 


19.  The  beam  is  a  cantilever  fixed  horizontally  in  the  wall. 
It  is  supposed  to  be  of  uniform  cross  section  and  of  weight  w  per 
unit  of  length,  and  to  have  a  weight  P  at  its  extremity. 

Y 


LINEAR  EQUATIONS  WITH  CONSTANT  COEFFICIENTS   61 


20.  The  beam  is  vertical.  It  has  rounded 
ends.  It  is  supposed  weightless.  It  is  de- 
flected a  small  amount  a  and  a  load  P  is 
applied  at  its  upper  end  just  sufficient  to  hold 
it  in  position. 


The  equation  for  a  circuit  containing  re- 
sistance, induction  and  capacity  in  terms  ot 
the  current  i  is 

n  t  r?  /7i*  i  \     fiP 

t*  I/         j.1/  u/if  t  JL    t*c 


in  terms  of  the  quantity  of  charge  q  is 

d?q       Rdq       jf_ 
~ 


where  e  denotes  the  E.M.F.,  R  the  resistance,  L  the  induction, 
C  the  capacity,  and  t  the  time  during  which  the  circuit  is  in 
operation.  In  each  of  the  three  following  exercises,  determine 
the  current  and  quantity  of  charge  in  the  circuit  after  a  time  t, 
supposing  that  the  resistance,  induction  and  capacity  are  constant. 

21.  The  E.M.F.  is  equal  to/(<).     Solve  when  R2C^  4L. 

22.  The  E.M.F.  is  constant  and  equal  to  E. 

23.  The   E.M.F.    is   a   simple   sine   function  of  the   time, 
=  E  sin  o)t.     Solve  when  R2  C  ^  4L. 

ANSWERS 


2    y  =  lx* 

3-  2/  =  ^±_2 
4.  y  =  \e*  + 


(8— Vl 


62     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

5.  y  = 


* 


7.  y  =  #(&?  +  el  +  eye).  8.  i/  =  z  +  2 

n  a;  sin  x 

9.  y  =  —  ^  —  -f  Cj  cos  a;  +  c2  sin  ^. 


10.  y  =  -  Jz  +  J-  +  c^-* 


11.     I/    =     —   -  ^  -    +  Cl  COS  ^ 

12.  y  =  J(cos  x  —  a;  sin  x  —  x  cos  x)  +  Cj  sin  ^  +  c2  cos  x  -f 

13.  y  =  <?(%a?  +  CjO?  +  cp  +  c3). 

14.  y  =  —  x*  —  24  +  Cjg*  -[-•  CjC"38  -f  c3  sin  x  +  c4  cos  #. 

P7  P  1    P78 

15.  4£Ty  =  ^a^  -  %-ar1.     Deflection  =7^^. 

Z  o 

U.SEI,  =  "-*<  -. 


1  P/8        5 
Deflects  =      - 


P  1  P/s 

18.  2£Iy  =  -  Fbf  +  ^x3.     Deflection  =  ™. 
o 


19. 


20    a- 

ZU.  «  =      -- 


Deflection  =         ^3  + 


Deflection  =  a  vers  \/^T-^-,  and  .  * .  P  = 


LINEAR  EQUATIONS  WITH  CONSTANT  COEFFICIENTS   63 

21  *'  = 


where 

T1  =  _  ,Mg  and 

JRC-  ^K*C2  -4LC 


The  value  for  <?  differs  from  that  for  i  only  in  having  /(£)  instead 
of  /'(<)•  _ 

RC- 


22.  1='^"          ^xc?  +c2e 


2LC  ~2LC~ 

when  R*G  <  4L. 


_ 

=  CE+cle 


cos 


R  t 
2L  sin 


2£C 

R  .  R  + 


. 

sm  a>£  + ^-q ^-2  cos  wit 


64     SHOET  COURSE  ON  DIFFERENTIAL  EQUATIONS 
where  T^  and  T2  have  the  values  given  in  exercise  21. 


q= 


JZV+        -i<«2 


CHAPTER  V 

HOMOGENEOUS  LINEAR  DIFFERENTIAL  EQUATIONS.     EXACT 
LINEAR  DIFFERENTIAL  EQUATIONS 

HOMOGENEOUS  LINEAR  DIFFERENTIAL  EQUATIONS 

44.  Definition.     A  homogeneous  linear  differential  equation  is 
an  equation  of  the  form 

dy 


where  pv  pv  •  •  •  ,  pn_v  pn  are  constants,  and  X  is  a  function  of 
x  but  not  of  y. 

This  equation  can  be  transformed  into  an  ordinary  linear  dif- 
ferential equation  with  constant  coefficients  by  changing  the  inde- 
pendent variable  from  x  to  2,  the  equation  of  transformation  be- 
ing x  =  ef.  The  equation  that  results  from  the  transformation 
may  be  solved  by  the  methods  of  the  last  chapter.  If  a  solu- 
tion is  y  =/(z),  the  corresponding  solution  of  the  original  equa- 
tion is  y  =  /(log  x). 

45.  The  transformation  and  general  solution  of  a  homogene- 
ous linear  differential  equation  in  the  general  case  will  not  be 
considered  here.  We  shall  merely  consider  them  in  a  particular 
example. 

EXAMPLE.     Find  the  general  solution  of  the  equation 
.  d*y  ,  dzy       _    dy 


Let  x  =  tf.     .'  .  z  =  log  x, 
dy      dy  dz       1  dy 
dx~~  dz  dx~  x  dz' 


d*y       d   /I  dy\  1   dy      1  d?y  dz       1   /dsy      dy\ 

dz*      dx  \  x  dz  I  x*  dz       x  dz2  dx      x*  y  dz2       dz  J 9 

65 


t 


66     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

d?y_(L\l_(<?y    ^M_I/^    «^,9 

<fe»~  dx\s<?\dz'~  dz)\  ~  x*\dz3  ~      dz*  "" 

•••*#  =  ?• 
dx      dz 

.#y_<Py     dy 

dx2  ~  dz*  ~  dz 

d*y      d*y       »dzy          dy 
^dx*  =  ~dz*~   6dz*+'dz' 

Substitute  in  the  equation.. 

d?y      d*y       .  dy 
• '  •  -rl  +  -y^  —  4  -f  _  4y  =  2. 
dz*  T  dz*          dz 

The  general  solution  of  this  equation   can   be  found   by  the 
methods  of  the  last  chapter.     It  is 

y  =  ~  fc  +  t  +  of*  +  c,e-2z  +  c^z. 
The  general  solution  of  the  original  equation  is  therefore 


y  =  -  J  log  x  +  I  +  ^  +  £  +  c3z2. 

EXACT  LINEAR  DIFFERENTIAL  EQUATIONS 
46.  Definition.     A  linear  differential  equation 


is  said  to  be  exact,  when,  if  the  left  hand  member  be  represented 
by  V,  the  expression  Vdx  is  the  exact  differential  of  some  func- 
tion U  which  does  not  contain  an  integral  of  y. 

The  expression  U  is  evidently  an  expression  actually  contain- 
ing a  derivative  of  order  n  —  1. 

47.  To  find  the  necessary  and  sufficient  condition  that  the 
equation  of  the  preceding  article  be  exact,  and  a  method  of  solu- 
tion of  such  an  equation. 


EXACT  LINEAR  DIFFERENTIAL  EQUATIONS         67 
Multiply  each  term  by  dx  and  take  the  integral  of  each  term. 


Now 

=fPnydy  identically. 


And,  by  integration  by  parts, 


where  the  primes  denote  differentiation  with  respect  to  x. 

.  •  .  fXdx  +  e  =  /(P.  _  P'n_t  +  P»B_a  _  P»'B_3  +  -  -  -  )ydx. 


Write  the  expression  in  brackets  as  Qn,  <?„_,,  •  •  •  ,  Qo  respec- 
tively. 

e  =fQnydz+Qn_iy+Qn_2y'+-  •  -  +  §0^.    (1) 

Now  in  order  that  the  equation  be  integrable  there  must  be  no 
term  in  the  right  hand  member  of  (1)  containing  an  integral  of 
y.  The  necessary  and  sufficient  condition  for  this  is  that  Qn  =  0. 


68     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

Therefore  the  necessary  and  sufficient  condition  that  the  equation 
be  exact  is  that  Qn  =  0  or 

P.  -  P'-l  +  P"»-,  -  ^"n-S  +  '  '  '   =  0.  (2) 

When  this  condition  is  satisfied  the  equation  reduces  to 

e.       (3) 


,  If  the  coefficients  in  (3)  satisfy  a  relation  in  Q  similar  to  (2) 
in  P,  equation  (3)  is  exact  and  the  above  process  may  be  re- 
peated. 

EXAMPLE.     Solve  the  equation 

.(l-*)+  (l-<W)-2,      +  a   =  6*.       . 


Here  P.  =  2,  _  P'n_t  =  2,  P"n_,  =  -  10,  -  P'"n_,  =  6. 
.'.  P.  -P^  +  PV,  -•*""-*  =  0. 

The  necessary  and  sufficient  condition  that  the  equation  be  exact 
is  therefore  satisfied. 

x-6x  ==  2x,      Qn_,  =  l_5^2 


The  equation  therefore  reduces  to 

A  VIA. 

dx*  dx 

In  this  equation,  Pn  =  2x,  —  P'n_j  =  4x,  P"n_2  -  —  6x. 
.  • .  Pn  _  P'      +  P"n     =  0. 

TI  n — 1     i  n — 2 

The  necessary  and  sufficient  condition  that  this  equation  be  exact 
is  therefore  satisfied. 

The  equation  therefore  reduces  to 


EXACT  LINEAR   DIFFERENTIAL  EQUATIONS        69 

This  equation  is  not  exact.  It  is  however  an  ordinary  linear  dif- 
ferential equation  of  the  first  order,  and  can  therefore  be  solved 
by  the  method  of  Art.  21.  The  general  solution  is 

y  =  -  i^log  (1  -  O  +  I x log  1  X_^ 


which  is  therefore  the  general  solution  of  the  original  equation. 

EXERCISES 
Find  the  general  solution  of  each  of  the  following  equations. 


. 

dx*  dx 

3.  ,»_3 

2 


dx 


2  -=4  —  4a;  ~  +  4i/  =  log  x. 
dx2  dx^ 


6    x->-2  2x  7    ^  x       +  v      loex 

~dx*+     dx=  '•      &?~     dx  +  y         g 

8.   (*-l)g  +  4.g  +  *  =  2* 

9.  (i+^g+4,|+2,=,.  • 

.  _      C?2^  C?l/ 

10.  -r^  —  cot  a;  -     +  cosec2  ^y  =  cos  a;. 


70     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

12.  O  +  3*')  g  +  2(1  +  6*)  g  +  6y  =  sin  * 

13.  (>3+  zs-  3*  +  1)       +  (9Z*  +  6*  -  9) 


dx*  dx*         dx  x2' 

ANSWERS 

1.   y  =  Klog  z)3  -  Klog  *02  +  If  ^g  x  +  c,  +  ^  +  ^-. 

u/        »t/ 

211  i  o*   i    />  /}»^    i      2 

.    i/  =  —  -%x  -\-  ^x  -j-  ^ . 

3.   i/  =  ^3  log  x  —  f  #3  -f  c^  -+  c2rc2  -f  c3#3. 

w    ^r:   T  -lOg  X  -4—  2"  — J—   C*X   — |—   Crt»t/     -y-       o~* 

O.    V  ^~  —  ^^  —  •'•^ff  *^  4~  @aX»  O.    V  ^~  'A'iC    -4—  —  -J—  C». 

7.  y  =  log  #  +  2  -f-  Cj#  log  a;  +  c^x. 


•  y  = 


10.  i/  =  ^  sin  ^  +  cx  sin  x  log  (cosec  #  —  cot  #)  4-  c2  sin  x. 

11.  = 


12.  (a;  +  3a;2)2/  =  —  sin  x  +  c^  -f  c2. 

13.  (#3  +  x*  —  3x  -f  l)i/  = 

2 


CHAPTER  VI 

CERTAIN  PARTICULAR  FORMS   OF  EQUATIONS 

dnv 

48.  An  equation  in  the  form  -=-j[  =/(#). 

case 

An  equation  in  this  form  is  exact  and  can  therefore  be  inte- 
grated by  the  methods  of  the  preceding  chapter.  It  can  also  be 
integrated  by  direct  integration. 

The  first  integration  gives 


where  ax  is  an  arbitrary  constant. 
The  second  gives 


where  a2  is  an  arbitrary  constant. 
After  n  integrations  there  results 

y  =  fff  •  • 

where  cv  ca,  •  •  •  ,cn  are  arbitrary  constants. 

49.  An  equation  in  the  form  -~  =/(y). 

dx 

An  equation  in  this  form  can  in  general  be  integrated  only 
when  n  =  1  and  n  =  2. 

When  n  =  1  the  equation  is 


71 


72     SHORT  COURSE  ON  DIFFERENTIAL   EQUATIONS 
To  integrate,  separate  the  variables. 


/dy 
ffi 


When  n  =  2  the  equation  is  in  the  form 


To  integrate,  multiply  by  2  -. 


••fed*'-'    ™'dx 
Now 

da;  d#*  ""  dx  \  dx 

d_ 

'  dx  \dx 


)'• 


Suppose  that  2j/(y)dy  =  ^(y). 


50.  An  equation  that  does  not  contain  #  directly. 
Such  an  equation  is  of  the  form 


CERTAIN  PARTICULAR  FORMS  OF  EQUATIONS      73 
Let 


Then 

d*y      dp      dp  dy          dp  ^ 

dx2  ~~  dx  ~  dydx  ~  P  dy  ' 


^y__^(^y\      *.(nfy\     A/«*\^ 

da?  ~  dx  \dx2}  ="  dx  \P  dy  I  ==  dy  \P  dy  )  dx 


and  so  on. 

The  equation  then  becomes  a  differential  equation  in  p  and  y 
of  order  n  —  1.  Suppose  that  it  can  be  solved  and  that  the  solu- 
tion is^>  =/(i/).  Then  a  solution  of  the  original  equation  is 

/dy 
m 

51.  An  equation  which  does  not  contain  y  directly. 
Such  an  equation  is  of  the  form 


dx9        '  d. 
Let 

dx 

The  equation  then  becomes  a  differential  equation  in  p  and  x 
of  order  n  —  1.  If  the  equation  can  be  solved  for  p  and  the 
solution  is  p  =  /(#),  a  solution  of  the  original  equation  is 


y  +  c  = , 

52.  An  equation  of  the  first  order  solvable  for  y. 

In  such  a  case,  when  solved  for  y,  the  equation  becomes 

y  =  F(x,p).  (1) 

6 


74       SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 
Differentiate  with  respect  to  x. 


This  equation  does  not  contain  y  explicitly.  It  is  an  equation  of 
the  first  order  in  p  and  x.  If  it  can  be  integrated  as  an  equation 
in  p  and  x,  there  results  on  integration  an  equation  between  x,  p 
and  an  arbitrary  constant.  From  the  resulting  equation  and 
(1),  if  p  can  be  eliminated,  there  results  an  equation  between 
x,  y  and  an  arbitrary  constant,  which  will  be  the  general  solution 
of  the  equation. 

53.  An  equation  of  the  first  order  solvable  for  x. 

In  such  a  case,    when  solved  for  x,    the  equation  becomes 
y  =  F(y,  p).     Differentiate  with  respect  to  y. 


The  method  of  procedure  from  this  point  is  similar  to  that  in  the 
preceding  article. 

EXERCISES 

Find  the  general  solution  of  each  of  the  twelve  following 
equations. 

<iny  o  d*y 

1.  -r€  =  x.  2.  -~  =  cos  x. 

dxn  dx* 


7 
• 


CERTAIN  PARTICULAR  FORMS  OF  EQUATIONS      75 


?  +  8      =  8*  10.  *  =       + 

dx*          dx  dx 

dy\*  dy 


13.  Find  the  curve  whose  curvature  is  constant  and  equal 
to  K. 

14.  If  a  sphere  of  radius  JKX  is  surrounded  by  a  concentric 
shell  of  radii  R^  and  Ey   the  potential  function,  V,  at  a  point 
either  in  the  space  between  the  conductors  or  outside  the  outer, 
satisfies  the  equation 


_ 

dr*  +  r  dr  "  U' 

where  r  is  the  distance  of  the  point  from  the  center  of  the  sphere. 

Solve  the  equation  given  that  Vl  is  the  potential  on  the  sphere 
and  V2  on  the  spherical  shell. 

15.  If  a  circular  cylinder  of  radius  ^  is  surrounded  by  a  cir- 
cular cylindrical  shell  of  radii  E3  and  jR3,  both  of  very  great 
length,  the  potential  function,  V,  in  the  space  between  the  con- 
ductors, is  such  that 


dr*  +  r  dr  ~    ' 

where  r  is  the  distance  from  the  point  to  the  axis  of  the  cylinder. 
Solve  the  equation  given  that  Vl  is  the  potential  on  the  cylin- 
der and  F2  on  the  spherical  shell. 

ANSWERS 


2.  y  =  —  cos  x  -j-  c^  +  c2. 

3.  y  =  —  log  x  —  J(log  #)2  +  <VB  +  cr 

4.  y  =  cx  sin  (ao;  -|-  c2). 


76     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

<Jy  ^ 

5.    =t  »  +  c,  =  — -  Vcx2/  —  2  +  — ,  log  (  V^  +  ^y  —  2). 


6.  y  =  c,ef*. 

I  ____^ 

7.  d=  a  +  cx  =  -  log  (cy  +  Vcy  —  1). 


9.  y  =  |  of  +      +  ca.  10. 


ll.  \x  +  cx  =  ^^x  —  y  —  log  (1  =p  Va?  —  y). 
c2. 

14    F       ^^     F.-F      Jg^ 
~ 


12.  y  =  ex  —  c2.  13.  A  circle  of  radius  -. 

K 


CHAPTER  VII 

ORDINARY  DIFFERENTIAL  EQUATIONS  IN  TWO  DEPENDENT 
VARIABLES 

54.  So  far,  the  differential  equations  considered  consisted  of 
two  variables,  one  independent  and  one  dependent.     We  shall 
now  consider  equations  in  three  variables.     These  may  be  divided 
into  two  classes  :  those  in  which  there  is  only  one  independent 
variable,  and  those  in  which  there  is  only  one  dependent  vari- 
able.    The  first  comes  under  the  class  called  ordinary  or  total 
differential  equations  :  the  second,  partial  differential  equations. 
This  chapter  is  taken  up  with  a  discussion  of  a  few  forms  of  ordi- 
nary differential   equations.      The  next  chapter  is  devoted  to 
partial  differential  equations. 

55.  If  /(#,  y)  is  a  single  valued  and  continuous  function  of 
the  two  independent  variables  x  and  y,  given  by  the  equation 

z  =  f(x,  t/),  and  ^-and  ^-are  continuous,  then,  by  definition, 

c/x         oy 

j        dz  ,        dz  1 
dz  =  —  ax  +  —  ay, 
dx          dy   * 

or 

dz  =  ^Hd  x  +  8J^dy.  (1) 

dx  dy 

If  f(x,  y,  z)  is  a  single  valued  and  continuous  function  of  the 
three  independent  variables  x,  y  and  zt  given  by  the  equation 

u  =/(#,  y,  z),  and  ^-,  —  and  -^-  are  continuous,  then,  by  defi- 

ux    <uy         oz 

nition, 


dz.    (2) 

dx  dy  dz 


77 


78     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

56.  Equation  (1)  of  the  preceding  article  has,,  as  a  special 
case  when  z  =  0,  the  equation 


=    . 
dx  dy 

That  is,  the  equation  is  true  for  the  equation  f(x,  2/)  =  0  where 
x  and  y  are  independent  variables.  If  y  in  /(#,  y*)  =  0  is  a 
single  valued  and  continuous  function  of  x,  the  equation  holds 
true  for  all  values  of  x  for  which  y  is  a  single  valued  and  con- 
tinuous function,  for  in  this  case  y  is  merely  restricted  to  values 
which  it  could  assume,  a's  well  as  others,  in  the  more  general 
case  where  it  is  independent. 

This  can  be  seen  more  clearly  perhaps  by  a  consideration  of 
the  geometrical  representations  of  the  equations.  ' 

The  equation  z  =  f(x,  y)  when  x  and  y  are  independent  vari- 
ables represents  a  surface.  If  z  =  0,  the  surface  is  the  ^-plane, 
and  the  equation 

df(x,  v)  7        df(X  T/)  7        A 
J\'  y}  dx  4-   J  V       dy  =  0 

dx  dy 

holds  true  for  every  point  in  the  plane.  If  y  is  a  single  valued 
and  continuous  function  of  x,  the  equation  /(#,  y)  =  0  repre- 
sents a  curve  in  the  xy-plane  in  which  the  equation  expressed  in 
the  form  y  =  <£(#)  gives  a  single  valued  and  continuous  function 
of  x,  and  since 

df(x,  y)  7        df(x.  y)  7 
J  V        da?  -f  —^-*2  dy  =  0 


holds  true  for  all  sets  of  values  of  x  and  i/  in  the  plane,  it  holds 
true  for  all  sets  of  values  which  together  determine  a  point  on 
the  curve  in  the  plane. 

57.  Equation  (2)  of  Art.  55  has  as  a  special  case  when  2  =  0, 
the  equation 

df(x.  y,  2)  7        df(x,  y,  z)  7        df(x9  y,  2)  ,        A 
JV  'y)  —  '-  dx  4-    t/v  '  y'  —  -  dy  4-    J^  'y  —  -  dz  =  0. 

dx  dy  dz 


EQUATIONS  IN  TWO  DEPENDENT   VARIABLES      79 

By  reasoning  similar  to  that  employed  in  the  preceding  article 
in  the  case  of  two  dependent  variables,  it  may  be  seen  that 
this  equation  holds  true  when  z  is  a  single  valued  and  continuous 
function  of  x  and  y. 

58.  An  integral  relation  in  x,  y  and  z,  equated  to  an  arbitrary 
constant  e,  say  <£(#,  y,  z)  =  c,  can  always  be  expressed  in  the 
form 

Pdx  +  Qdy  +  Rdz  =  0, 

where  P,  Q  and  R  are  functions  of  x,  y  and  z,  and  do  not  con- 
tain the  arbitrary  constant  c. 

For,  the  result  of  taking  the  differential  of  each  member  of 
the  equation  <f>(x,  y,  z)  =  c  is,  by  the  preceding  article, 


y,  0  dx    a*(*  y,  0  d     a»(*  y,  0  dg  =  0 

dx  dy  dz 

and  this  equation  is  in  the  specified  form. 

EXAMPLE.    The  result  of  taking  the  differential  of  each  mem- 
ber of  the  equation  x2y  —  xz*  -f-  y*z  =  c  where  c  is  arbitrary,  is 

(2xy  _  z^dx  -f  (x2  +  '2yz)dy  +  (y2  —  2a«)cfe  =  0. 

This  equation  is  in  the  form  Pdx  -f-  Qdy  -f  .Kcfe  =  0. 

The  resulting  equation  Pdx  -f  Qdy  -f  Ifcfe  =  0  is  such  that  P, 
Q  and  R  are  proportional  to 

d<f>  d$  ,  d<£ 
—  ,  —  and  •=-, 
dar  ^t/  ^s 

respectively. 

Conversely,  however,  an  equation  of  the  form 


Pdx  +  Qdy  +  J2dz  =  0 

where  P,  Q  and  JR  are  functions  of  x,  y  and  2,  does  not  neces- 
sarily give  rise  to  a  solution  of  the  form  <j>(x,  y,  z)  =  c.  This 
can  be  seen  immediately  because  an  equation  of  the  form 

Pdx  +  Qdy  -|-  Rdz  =  0 


80     SHORT  COURSE  ON   DIFFERENTIAL  EQUATIONS 

which  gives  rise  to  a  relation  <£(#,  y,  2)  =  c  must  be  such  that 
P,  Q  and  R  are  proportional  to 

a*      a*     and     a* 

ai'  ^  a^' 

respectively,  and  these  relations  cannot  hold  for  all  values  of  P, 
Q  and  R. 

59.  To  determine  when  an  equation  of  the  form 
Pdx  +  Qdy  +  Rdz  =  0 

has  a  solution  of  the  form  <£(#,  y,  z)  =  c. 

If  it  be  assumed  that  Pdx  -f  Qdy  -\-Rdz  =  0  has  a  solution 
<£(#,  y,  z)  =  c,  then  P,  Q  and  P  must  be  proportional  to 


respectively,  or 


<  ,        < 

•—,     3^     and     ~, 

dx      dy  dz 


where  ft  is  a  certain  unknown  function.     From  the  first  two  of 
these  equations  there  results 


or 


(1) 


Similarly,  by  using  the  first  and  third  equations  we  get 


and  by  using  the  second  and  third, 

q    as 


,ON 


EQUATIONS  IN  TWO  DEPENDENT    VARIABLES      81 

Multiply  equations  (1),  (2)  and  (3)  by  R,  Q  and  P  respec- 
tively, and  add. 

Q      dR\          idR      dP\         /dP     dQ\ 

~  '~ 


Therefore,  if  the  equation  Pdx  +  Qdy  -\-  Rdz  =  0  has  a  solution 
<£(#,  y,  z)  —  c,  equation  (4)  must  be  satisfied. 

Conversely,  if  equation  (4)  is  satisfied,  the  equation 


Pdx  +  Qdy 

has  a  solution  <£(#,  y,  z)  =  c.  The  proof  of  this  theorem  is  some- 
what long  and  will  not  be  given  in  this  book.*  The  theorem 
however  will  be  assumed  in  the  subsequent  work. 

Definition.  Equation  (4)  is  called  the  condition  of  integra- 
bility  of  the  equation  Pdx  -f-  Qdy  -f  Rdz  =  0. 

60.  To  solve  the  equation  Pdx  -f-  Qdy  -f  Rdz  =  0  when  the 
condition  of  integrability  is  satisfied. 

Suppose  at  first  that  z  is  constant  so  that  the  equation  becomes 
Pdx  -j-  Qdy  =  0.  Solve  this  equation.  Suppose  that  the  solu- 
tion is  f(x,  y,  z)  =  a  constant.  Let  u  =f(x,  y,  z).  Find  a 
quantity  /x,  such  that 


du 


Multiply  the  equation  Pdx  -|-  Qdy  -f-  Rdz  =  0  by  ft. 
.  •  .  fji(Pdx  -f  Qdy  +  Rdz)  =  0. 

This  equation  may  be  written  in  the  form  du  -f-  Sdz  =  0  where 
u  and  S  are  in  general  functions  of  x,  y  and  z.  In  the  equation 
du  -j-  Sdz,  change  the  variables  from  x,  y  and  z  to  x,  u  and  z  by 
means  of  the  relation  u  —  f(x,  y,  z).  The  equation  then  be- 

*  For  a  proof  of  this  theorem  and  also  that  iS/  of  Art.  60  does  not  con- 
tain x,  the  student  is  referred  to  Forsyth,  A  Treatise  on  Differential  Equa- 
tions, Art.  152. 


82     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

comes  du  -f  S'dz  =  0.  It  can  be  shown  that  S'  does  not  contain 
x.  Assuming  that  it  does  not,  the  equation  du  -f-  S'dz  can  be 
integrated  as  an  equation  in  u  -and  z.  The  general  solution  of 
the  equation  is  the  general  solution  of  the  original  equation. 

EXAMPLE.     Solve  the  equation 

yz      j            xz      ,  _.y  1       A 

-=-2 — -  dx  —  -= , dy  —  tan  *-  dz  =  0. 

«H       I       ni*  />.*       I       fti*       **  n* 


Suppose  that  z  is  constant.     The  equation  then  becomes 


dx  —  -j — -^— ; j  dy  =  0     or    ydx  —  xdy  =  Q. 


s        2       —  -=  -  , 

&  +  y        *  +  f 

The  solution  of  this  equation  is 


x 

—  =  a  constant. 

y 


Let  u  =  -. 


Let  pP  =  -. 


+ 


£2  4-  t/2 

Multiply  the  original  equation  by  —  :~^-. 

y 

.'.~dx^--2dy^  Xl±£tKTiy-dz  =  0. 

y         y1  2/  z  * 

Now 

du  =  -dx  —  -=  dy. 

y        y 


,.___ 

fz  x 

Substitute 


x 


EQUATIONS  IN  TWO  DEPENDENT  VARIABLES      83 
in  this  equation,  y  being  derived  from  the  equation 


u 
Separate  the  variables. 


dz 


- 
u 


Let  tan  l  -  =  v. 


v        z 

.  '  .  VZ  =  C. 

.  • .  z  tan"1  -  =  c. 
x 

Therefore 

_ly 

x  ~~ 

is  the  general  solution  of  the  original  equation. 
61.  Suppose  that  in  the  equation 

Pdx  +  Qdy  +  Rdz  =  0 

the  condition  of  integrability  is  not  satisfied.  Then  there  is  no 
relation  <£(#,  y,  2)  =  c  which  satisfies  the  equation.  In  such  a 
case  a  relation 


is  assumed  arbitrarily  and  a  relation  <£(#,  yy  25)  =  c  is  sought 
which,  together  with  ^(#,  y,  2)  =  0,  will  satisfy  the  equation. 
By  differentiation  of  \f/(x,  y,  2)  =  0  there  results 


From  this  equation  and  (1)  suppose  that  z  and  dz  be  eliminated. 
Then  there  will  result  an  equation  of  the  form  P'dx  -f-  Q'dy  =  0 


84     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

where  P'  and  Q*  are  functions  of  x  and  y  the  values  of  which 
depend  upon  \j/(x,  y,  z).  Suppose  that  a  solution  of  this  equation 
containing  an  arbitrary  constant  is  found  and  is  <£(#,  yy  z)  =  c. 
Then  this  solution  and  \l/(x,  y,  2)  =  0  together  give  a  solution 
of  the  equation. 

As  an  illustration  consider  the  following  example  : 

The  equation 


considered  in  exercises  47  to  50  inclusive,  Chapter  III,  for  special 
cases  of  e,  does  not  satisfy  the  condition  of  integrability  if  e,  i 
and  t  are  variables  independent  of  each  other.  For,  the  equa- 
tion may  be  written  as 

Ldi  +  (Ri  —  e)dt  +  0  •  de  =  0. 
By  application  of  the  condition  of  integrability  there  results 


or 


Since  L  is  not  zero,  the  equation  does  not  satisfy  the  condition 
of  integrability.  Assume  e  =/(£),  however,  and  the  equation 
becomes  an  ordinary  linear  differential  equation  of  the  first  order. 
The  solution  is 


'•-r~:'/< 

JU  */ 


From  this  solution  the  results  of  exercises  47  to  50  inclusive, 
Chapter  III,  may  be  found  by  substitution. 

62.  The  cases  considered  thus  far  consisted  of  one  equation  in 
two  dependent  variables.  Another  important  class  of  equations 
is  the  case  of  two  total  differential  equations  in  two  dependent 


EQUATIONS  IN  TWO  DEPENDENT  VARIABLES      85 

variables  where  each  equation  is  of  the  first  degree  with  constant 
coefficients.  The  method  of  solution  of  this  class  of  equations  is 
as  follows  : 

By  differentiation  and  elimination,  obtain  one  equation  in  one 
unknown.  This  equation  may  be  solved  by  methods  previously 
discussed.  The  solution  found  must  be  a  solution  of  the  original 
equations.  Another  solution  is  found  by  substituting  the  one 
just  found  in  the  equations.  The  complete  solution  consists  of 
two  linearly  independent  relations  between  the  variables. 

EXAMPLE.     Solve  the  equations 

!-2y-M  =  0.  ...       (1) 

^  -  Qy  +  5z  =  0.  (2) 

Differentiate  (2)  with  respect  to  x. 

•  d'y    6^4-5*     o  m 

•&?-*dx  +  dd-x  =  "- 

Multiply  (1)  by  -  5,  and  add  to  (2)  and  (3). 

.•.g-5^  +  4y  =  0. 

dx*         dx T 

This  is  a  linear  differential  equation  of  the  second  order  with  con- 
stant coefficients  and  right  hand  member  zero.  It  can  therefore 
be  solved  by  the  methods  of  Art.  31. 

.'.y  =  c^  +  c2e*.  (4) 

Substitute  this  value  of  y  in  (2)  and  solve  for  z. 

.•.*  =  cle*  +  2fe<*.  (5) 

Equations  (4)  and  (5)  together  constitute  a  set  of  solutions  of 
the  given  equations. 


86     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

EXERCISES 

In  each  of  the  seven  following  equations,  show  that  the  condi- 
tion of  integrability  is  satisfied.     Solve  the  equation. 

1.  (y  +  z)dx  +(z  +  x)dy  +  (x  +  y^dz  =  0. 

2.  (2x>y  +  2xf  +  2xyz  +  l)da?  +  (x*  +  x2y  +  x2z  +  2xyz 

+  2fz  +  2yz*  +  l)dy  +  (xy*  +  f  +  tfz  +  1)&  =  0 

3.  (2xy  +  z^dx  +  <>2  +  2y»)dy  +  (y*  +  2xz)dz  =  0. 

4.  (a  -f  z)ydx  +  (a  +  z)xdy  —  xydz  =  0. 

5.  (y  +  *)(»  +  *)*c+(a;+a)(a+A)dy+(a?+a)(y+6)(fe=:0. 

6.  (yz  +  2x)dx  +  (xz  +  2y)dy  +  (xy  +  2z)dz  =  0. 

7.  (2xyz  +  fz  +  yz^dx  +  (x*z  +  2xyz  +  xz^dy 

+  (^  +  *f  +  2xyz)dz  =  0. 
Solve  the  following  sets  of  equations. 


10. 


EQUATIONS  IN  TWO  DEPENDENT  VARIABLES      87 
||  +  ,  +  8.  =  <r,     |  +  83,  +  4*  =  e*. 


1  Q       d_   __   zAl  ——   2z      '     f  • —      I      ^)7/  __   2z  —  f* 

ax  ct/x 

ANSWERS 

1.  xy  4-  yz  -\-  zx  =  c. 

2.  x*y  4-  y*z  +  log(#  +  y  +  z)  =  c. 

3.  x*y  4.  i/22  -f-  z*x  =  c.  4.  xy  =  c(a  -j-  3). 

5.   (#  -f-  a)  (y  4-  ^)  (2  +  ^)  =  c-     6.  ^1/3  4-  x2  4-  y2  4.  22  =  c. 
7.  #?/2(#  4-  y  4-  2)  =  c. 

*7^>  7^*  7/* 

Q  — jjj  — 3jg  I  v/j     — Sx  2      3x  t        — Sx 

10.  y  =  c^-635  4-  ef»,     z  =  3c}e^  -  c^x. 

11.  y  =  cj 

V7  V7 

12.  y  =  Cjg^  cos  -^  x  4-  c2et*  sin  -^  re, 


2 


C  1+/65 

~ 


88      SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

13.  y  =  c^e"*  cos  V&&  -f-  c.2e~x  sin  V&E, 

z  =  —  ct  V^e"*  sin  V5#  +  ca  V5e~*  cos  V&c. 

14.  y  =  —  x  —  ^  +  c^  +  c^-*, 
s  =  —  1  —  2o;  -  f  ef  - 

15.  y^A^ 

16.  y  =  ^631 


17.  y  =  -  y^  +  fa  +  V-^-yr^  +  c 


18.  y  =  T6*  —  i^2*  +  Cig5x  cos 


V5c 


19.  y  =  iVe31  +  |e*x  +  Cje2x  cos  VlO*  +  c,ete  sin 

z  =  -  T5Te3l+  |ete+^S 
1. 


CHAPTER  VIII 

PARTIAL  DIFFERENTIAL  EQUATIONS. 

63.  So  far  we  have  considered  differential  equations  in  which 
there  is  only  one  independent  variable.     We  shall  now  consider 
equations  involving  two  independent  variables.     Such  equations 
belong  to  the  class  called  partial  differential  equations, 

In  this  book,  the  independent  variables  will  be  denoted  by  x 
and  y,  and  the  dependent  variable  by  z.  The  partial  derivative 
of  z  with  respect  to  x  and  with  respect  to  y  will  be  denoted  by  p 
and  q,  respectively. 

Definition.  A  linear  partial  differential  equation  of  the  first 
order  is  an  equation  of  the  form 

Pp+Qq  =  R, 

where  P,  Q  and  E  are  functions  of  x,  y  and  z,  and  do  not  con- 
tain p  or  q. 

64.  If  there  are  two  equations  containing  x,  y  and  z,  p  and  q, 
which  can  be  solved  for  p  and  q,  the  result  may  be  substituted  in 

dz  =  pdx  -f  qdy 

thus  giving  an  ordinary  differential  equation.     Usually,  however, 
there  is  only  a  single  differential  equation  given. 

65.  Derivation  of  a  partial  differential  equation. 

(a)  By  the  elimination  of  constants.  Let  <#>(#,  y,  z,  cv  c2)  =  0 
be  a  relation  between  x,  y,  z  and  two  arbitrary  constants  ct  and 
c2.  By  differentiation  of  <£(#,  ?/,  z,  cv  c2)  =  0  with  respect  to  x 
holding  y  constant  there  results 


90     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

By  differentiation  with  respect  to  y  holding  x  constant  there 
results 


By  means  of  these  two  equations  and  <£(#,  y,  z,  cv  c2)  =  0,  cv 
and  c2  can  be  eliminated.     The  result  is  an  equation 

F(x9  y,  z,  p,  q)  =  0 
which  is  a  partial  differential  equation  of  the  first  order. 

EXAMPLE.  Let  a?  -f  y2  4-  z*  4-  c^x  4-  c2?/  =  0  be  an  equation 
between  x,  y  and  z,  and  two  arbitrary  constants  ct  and  c2.  By 
differentiation  with  respect  to  x  holding  y  constant  there  results 

2x  4-  cx  -f-  2zp  =  0. 

By  differentiation  with  respect  to  y  holding  x  constant  there 
results 

2y  4-  c2  4-  2zq  =  0. 

By  elimination  of  c,  and  c2  between  the  three  equations  there 
results 

x2  +  y*  —  z*  +  2#zp  4.  2ysg  =  0. 


This  is  a  partial  differential  equation  of  the  first  order. 

(6)  By  the  elimination  of  an  arbitrary  function.  Suppose 
that  u  and  v  are  functions  of  the  variables  x,  y  and  z,  and  that 
<l>(u,  v)  —  0  where  <f>(u,  v)  is  an  arbitrary  function  of  u  and  v. 

The  differential  of  $(u,  v)  =  0  is 


-  -       = 

du  dv 

Now 

,        du  .,        du 

du  =  —  a«  4-  -5- 

r 


when  y  is  constant,  and 

, 
d 

when  x  is  constant,  and  similarly  for  v. 


,          it  ,          w 

du  =  —  dy  +  — 
dy    '        dz 


PARTIAL  DIFFERENTIAL  EQUATIONS 


91 


Therefore  the  partial  derivatives  of  the  equation  <j>(u,  a;)  =  0 
with  respect  to  x  and  y,  respectively,  are 


and 


•  r  du      du 


'  du 

'  du      du 


d$dv      dv 


]_!__?    —  _j_  —  g     =00 
i—    y  j 


du\_dy  ^  dz 
Eliminate  w-  and  x— -  from  these  equations. 

UU  CV  i 

du 


When  arranged  in  powers  of  p  and  q  and  the  coefficients  ex- 
pressed as  determinants,  the  equation  becomes 


du  du 

du  du 

du  du 

dy  dz 

dz  dx 

dx  dy 

P  + 

q  = 

dv  dv 

dv  dv 

i 

dv  dv 

dy  dz 

dz  dx 

dx  dy 

This  may  be  written  in  the  form 


where 


(1) 


R 


du     du 

~~ 

du     du 

du     du 

dy     dz 

dz     dx 

dx     dy 

dv      dv 

dv      dv 

dv      dv 

dy      dz 

dz     dx 

dx      dy 

This  is  a  partial  differential  equation  of  the  first  order.  There- 
fore from  the  equation  </>(^,  v~)  =  0  a  partial  linear  differential 
equation  of  the  first  order  can  be  formed  which  does  not  contain 
the  arbitrary  function  <t>(uy  v). 

EXAMPLE.  Suppose  that  u  =  x  -f  y  -f  z  and  v  =  x*+  y*-\-  z*. 
Let  <f>(u,  v~)  =  0  be  an  equation  connecting  u  and  v  where 
<l>(u,  v)  is  an  arbitrary  function  of  u  and  v. 


92     &HORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

By  differentiation  of  $(u,  v)  =  0  with  respect  to  x  and  with 
respect  to  y  there  result 


and 


- 


respectively.     By  elimination  of  ^-  and  -^  from  these  equations 
there  results 


1     1 

3      y 


1      1 


1       1 

y    x 


or 


This  is  a  partial  linear  differential  equation  of  the  first  order 
which  does  not  contain  the  arbitrary  function 


66.  We  have  seen  that  a  differential  equation  with  two  inde- 
pendent variables  can  be  derived  from  an  expression  containing 
two  arbitrary  constants  or  from  an  expression  containing  an  arbi- 
trary function  of  two  independent  functions  of  the  variables.  We 
see  therefore  that  a  differential  equation  with  two  independent 
variables  may  involve  in  its  solutions,  arbitrary  constants  or  an 
arbitrary  function  of  the  variables. 

Definitions.  A  relation  between  the  variables  of  a  differential 
equation  with  two  independent  variables  which  includes  two  arbi- 
trary constants  is  called  a  complete  integral  of  the  equation. 

A  relation  between  the  variables  of  a  differential  equation  with 
two  independent  variables  which  involves  an  arbitrary  function 
of  two  independent  functions  of  these  variables  is  called  a  general 
integral  of  the  equation. 

There  is  another  class  of  solutions  called  singular  integrals  but 
these  will  not  be  considered  here. 


PARTIAL  DIFFERENTIAL  EQUATIONS 


93 


67.  Consider  the  two  equations  u  =  cx  and  v  =  c2  where  ^  and 
v  are  functions  of  #,  y  and  z,  and  cx  and  c2  are  arbitrary  con- 
stants. By  differentiation  of  u  =  el  and  v  =  c2,  there  result 

dw 


7 

dx  +    - 
* 


and 


7        ~ 

^-  =-  dz  =  0 

dy    '        dz 


dv  7 

^d- 

B  +  ^-  dy  +  5- 
dy           dz 

dz  =  0, 

respectively. 

Multiply  (1)  by^-, 

cz 

(2)  by  ^-  ,  and  subtract. 

du     du 

du 

du 

dz      dx 

dy 

'dz 

• 

dx  — 

dy  ^  0. 

dv     dv 

dv 

dv 

dz      dx 

dy 

~dz 

Multiply  (1)  by  ^-,  (2)  by  ^-  ,  and  subtract. 

du      du 

du 

du 

dx      dy 

dy 

dz 

• 

dx  — 

dz  :=  0. 

dv 

av 

dv 

dv 

dx~ 

aj 

dy 

dz 

dx 

dy 

dz 

du      du 

du     du 

du     du 

dy      dz 

dz      dx 

dx     dy 

dv      dv 

dv      dv 

dv     dv 

dy      dz 

dz      dx 

dx     dy 

(1) 

(2) 


Now  <£(w,  v)  =  0  is  a  general  integral  of  the  equation 


if 


Q 


R 


du   du 

du  du 

" 

du   du 

dy   dz 

dz   dx 

dx   dy 

dv   dv 

dv  ,dv 

dv   dv 

dy   dz 

dz   dx 

dx   dy 

See  Art.  65. 


94     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

Therefore   <£(X   v)  =  0  is  a  general  integral  of  the  equation 
Pp  +  Qq  =  R  if  u  =  cx  and  v  =  c2  are  solutions  of  the  equations 

dx       dy      dz 

p  =  (2  =  3T' 

68.  From  the  investigations  of  Arts.  65  and  67  the  following 
rule  for  finding  a  general  integral  of  the  linear  partial  differential 
equation  Pp  -j-  Qq  =  E  is  determined. 

Solve  the  equations 

dx      dy      dz 


Suppose  that  u  =  ct  and  v  —  c2  are  two  independent  integrals  of 
these  equations.  Then  $(u,  v)  =  0  where  <£(i£,  v)  is  an  arbi- 
trary function  of  ^  and  v  is  a  general  integral  of  the  equation 
Pp+Qq  =  B. 

Definition.     The  equations 

dx      dy      dz 
P  =  Q=K 

are  called  the  subsidiary  equations  of  Pp  +  Qq  =  E.  They  are 
also  sometimes  called  Lagrange's  equations. 

69.  As  illustrations  of  the  method  of  solution  of  a  linear  par- 
tial differential  equation  of  the  first  order,  consider  the  following 
examples. 

EXAMPLE  1.     So]ve  the  equation  x*p  +  xyq  +  y2  =  0. 
Write  the  subsidiary  equations 

dx      dy  dz 

x*  ~  xy~    "  i/2' 
Solve  the  equation 

dx      dy  x 

y        •  f> 

x*  ~  xy          "  y  ~ 
Solve  the  equation 

dy  dz 


PARTIAL  DIFFERENTIAL  EQUATIONS  95 

From 

x 

—  c» 

substitute  the  value  of  x,  and  the  equation  becomes 

dy 

-Z  =  —  dz. 


A  general  integral  of  the  original  equation  is  therefore 


EXAMPLE  2.     Solve  the  equation  (y  —  z)p  +  (z—x')q=x— -y. 
Write  the  subsidiary  equations 

dx  dy  dz 

From  a  familiar  theorem  of  algebra,  if 
ace 

i  =  d=f 

then  la  +  me  -|-  ne  =  Ib  -f  md  -f  nf  where  /,  m  and  n  are  any 
multipliers  whatsoever.  Application  of  this  theorem  to  the  sub- 
sidiary equations  gives 

dx  4-  dy  -f  dz  =  0,  (1) 

when  I  =  m  =  n,  and 

xdx  -\-  ydy  -f-  zdz  =  0,  (2) 

when  I  =  x,  m  =  y,  n  =  z. 

Solve  equations  (1)  and  (2).  Therefore  x  -\-  y  -\-  z  =  ^  and 
x2  -f-  i/2  -}-  22  =  ct  are  solutions  of  equations  (1)  and  (2),  and 
therefore  of  the  subsidiary  equations.  A  general  integral  of  the 
original  equation  is  therefore  <f>(x  +  y  -|-  z,  x*  -f-  y2  -f  22)  =0. 


96     SHORT  COURSE  ON  DIFFERENTIAL   EQUATION 

EXERCISES 

Determine  the  partial  differential  equations  of  which  the  four 
following  equations  are  complete  solutions,  cx  and  c2  being  arbi* 
trary  constants. 

x  +  c$.  2.  z2  =  Cj#2  -f  c2i/2. 

*  +  oCy  +  ',)•          4.  £  +  £+£=  1. 

ci         ca 

Eliminate  the  arbitrary  function  from  each  of  the  four  follow- 
ing equations. 

5.  <f>(x+y-z,  a?+y*-s?)=0.      6.   <j>(x  +  y  +  z,  z)  =  0. 

7.  *  =  ety(s  +  y).  8.   z=S(*  +  ft. 

Find  a  general  integral  of  each  of  the  following  equations. 

9.  xzp  —  yzq  =  xy.  10.  x*p  -\-  y*q  —  z2  =  0. 

11.  x*yp  -f-  yq  =  a;2z.  12,  xp  —  yq  =z  x  —  y. 

13.  (^  _  z2)^  +  (,2  -  o;2)^  +  (i/2  -  ^2)  =  0. 

14.  (2z  -  3y)i>  +  (Sx  - 


ANSWERS 

1.  xp  +  yq  =  z.  2.  xp  +  yq  =  z. 

3.  j9£  =  2-  4.  a;zjo  +  2/zg  —  z2  -|-  a2  =  0. 

5.    (y—  z)p  +  (z—  x)q=y—x.        6.  p  —  q  =  0. 

7.  p  —  q  =  —z.  8.  yp  —  xq  =  0. 

10. 


.,~=.      .        -,- 

11.  <#>  |  v  +  -,  ;  -  ^r-«  log  ^  4-  -  r  —  log  z  \  =0. 
^  r  r  x9  (xy  +  I)2  r  xy  +  I          >    J 

12.  <#>(^,  a?  +  y  -  «)  =  0. 

13.  <K*  +  y  +  z,  of  +  f  +  z3)  =  0. 

14.  <K4*  +  2y  +  3z,  ^  +  y2  +  z2)  «  0. 


CHAPTER  IX 

APPLICATIONS   OF  PARTIAL   DIFFERENTIAL   EQUATIONS. 
INTEGRATION   IN   SERIES 

APPLICATIONS  OF  PARTIAL  DIFFERENTIAL  EQUATIONS 

70.  It  is  shown  in  the  Analytical  Theory  of  Heat  that  the 
change  of  temperature  in  any  solid  at  a  point  (#,  y,  2)  within 
the  solid  is  given  by  the  equation 

du       2  F  d*u       d*u      d2u~\ 
=G  +        +          ' 


l.  _  C2    _Ji  i  — _  i      _*1  i  _._?*  .  C3) 

5  [  cV  ^  r   dr  ^  r2  d<£2  ^  cV 


where  u  represents  the  temperature  at  the  point  and  t  denotes 
time. 

In  polar  or  spherical  coordinates  the  equation  becomes 

_  /  „  du  \  ->/•/»  du N 

9     d  \  r  ^-  ]  d  [  sin  o 

cm       c          \     dr  /  \ 

dt  ~~  r2  dr  sin  ^  ^  r  sin2 

and  in  cylindrical  coordinates, 

du 

'dt 

If  the  solid  is  a  rectangular  plate  so  thin  that  the  thickness 
need  not  be  taken  into  account,  equation  (1)  becomes 

du 

~dt 

If  the  solid  is  a  wire  of  infinite  extent  so  thin  that  the  breadth 
or  thickness  need  not  be  taken  into  account,  equation  (1) 
becomes 

du        zd*u  . 

In  the  case  of  a  sphere  when  the  temperature  u  depends  merely 
on  the  distance  of  the  point  from  the  center,  equation  (1),  as 
8  97       -   , 


98     SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 
can  be  seen  from  (2),  reduces  to 


= 

dt  dr* 

In  the  problem  of  permanent  states  of  temperature,  du/dt  =  Q, 
and  the  equation  becomes 


_ 

dtf        df        dzz 

an  equation  known  as  Laplace's  Equation,  and  sometimes  writ- 
ten V2u  =  0.  This  equation  also  figures  in  the  Theory  of 
Potential. 

In  polar  or  spherical  coordinates,  equation  (7)  becomes  the 
right-hand  member  of  equation  (2)  set  equal  to  zero,  and  in 
cylindrical  coordinates,  it  becomes  the  right-hand  member  of 
(3)  set  equal  to  zero. 

In  the  Theory  of  Acoustics,  in  considering  for  instance  the 
transverse  vibrations  of  a  stretched  elastic  string,  there  occurs 
the  equation 


and   if  the  resistance  of  the  air  be  taken  into  account,   the 
equation 


In  the  problem  of  the  vibrations  of  a  stretched  elastic  mem- 
brane, there  occurs  the  equation 


(10) 

~di'=          '          ' 

which  in  cylindrical  coordinates  becomes 


df  = 

71.  As  an  illustration  of  transformation  of  coordinates,  con- 
sider the  transformation  of  Laplace's  Equation  in  two  dimen- 
sions, 


APPLICATIONS  OF  EQUATIONS  99 

&u      d*u  _ 
d~x>  +  'df  ~     ' 

from  rectangular  to  polar  coordinates. 

The  equations  of  transformation  are  x  =  r  cos  0,  y  =  r  sin  0. 

Now  u  is  a  function  of  x  and  y  and  therefore  of  r  and  0. 
Therefore,  as  seen  in  calculus, 

7        du  .        du  7A 


If  y  is  held  constant,  this  equation  becomes, 
du  7         du  i  « 


Divide  by  A#,  or  what  is  the  same,  dx,  and  there  results  ths 
equation 

du       du  dr      du  dO  . 

~dx=~dr  dx  +  dO  d^c°  W 

Similarly, 

du      du  dr      du  dO 

dy  =  dr  fry  +  dO  dy 

Since  x  =  r  cos  0  and  y  =  r  sin  9,  therefore  r  =  V#2  +  y*  and 
0  =  tan"1  y/x. 

dr  x  x 


and 
Now 
and 

'  dx~ 
dr 

&+t 

r 

y 

sin  0 

dy~ 
dO 

y 

y            sin0 

dO 

x*  +  y 
x 

ty&                               n* 

x       cos  0 

dy 
dx  ~  dr  dx 

/v.2        I        „.' 

du  dO 
+  dOdx 

'-r2-     r    • 

du        -       du  sin  0 
=  drC(        ~d0  ~7~' 

100    SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

dzu       d  [du       ^     du  sin  0~|        A      d  [du  du  sin  01  sin  b 

^j«  =  5"  \^T  cos0—  ^-  -         cos0—  ^A  7T-COS0--  ^~ 

dx*       dr\dr  dO    r    J  dO^dr  dO     r     \    r 

[d*u        .        d*u    sin  0      du  sin  01 
—  -  cos  0  —  ^—  ^  --  +  ^  —  p- 
^r2  ^r^0     r      ^  ^0     r2  J 

[^tt  .    A 
-d-rsme 


cos  0 
d*u  sin  0      du  cos  01  sin  0 


r 
Similarly 

d*u       f  d*u  .    ..        d*u    cos  0      du  cos  0 1    , 
^-2  =     ^  sin  0  +  ^-^  -    -  —  3Tn  — r     sm  0 

Cy  \Or  CTCU       T  Cv       T 

du        n      d2u  cos  0      du  sin  0 1  cos  0 
a0"  "T~   "  dO  ~V~     ~r~  * 


^a:2       di/2       ^r2       r  ^r      r2  d(P* 
Therefore  the  equation 

dtf  +dy*=    ' 
in  polar  becomes 

d*u      1  du      1   d2^      .. 

* 

72.  A  method  of  determining  particular  solutions  of  those  of 
the  above  equations  with  constant  coefficients  is  illustrated  in  the 
following  example. 

EXAMPLE.     Find  particular  solutions  of  the  equation 

— 2  =  °2 

Assume   that   there   is   a   particular   solution    in    the    form 
z  =  e^+Pv+y*  where  a,  ft  and  y,  are  constants. 
Substitute  in  the  equation. 


Now  e^+Py+v*  cannot  be  zero  for  any  values  of  x,  y  and  t. 


APPLICATIONS  OF  EQUATIONS  101 

Therefore  z  =  e«*+Py*cy  «*+(**'  is  a  particular  solution  of  the 
equation  where  a  and  ft  are  arbitrary  constants. 
,     The  above  solution  can  be  put  into  another  form  as  follows: 
Let  a  =  aj  and  fi  =  fij  where  j  =  V  —  1. 
Then 

z  =  eOan-Py*0*1'*2  +£*)./. 
Therefore 

z  =  sin  (ax  +  fiy  ±  ct  Va2  +  /32),  (1) 

and 

^  =  cos  (oa?  +  jfy  ±  ct  Va2  +  £2),     (See  Art.  5.  )     (2) 

are  particular  solutions  of  the  equation. 

\    From  these  can  be  found  particular  solutions  in  the  forms 


z  =  sin  ax  sin  /3y  sin  ct  Va2  +  /32, 
z  =  sin  cue  sin  fiy  cos  c£  V«2  +  ft2, 

and  six  others.     The  determination  of  these  six  is  left  as  an 
exercise  to  the  student. 

73.  Consider  the  equation 


~^7~       551  "  ~dO~ 

which  is  Laplace's  Equation  in  spherical  coordinates  where  u  is 
independent  of  <£. 

Let  u  =  rmP,  where  P  is  a  function  of  0  alone,  and  m  is  a 
positive  integer.     On  substitution  there  results  the  equation 


sin  v  cv 

Change  the  independent  variable  from  0  to  x  where  x  =  cos  0. 


The  solutions  of  equation  (1)  are  known  when  P  is  determined 
from  equation  (2).     Equation  (2),  not  only  when  m  is  a  posi- 


102    SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

live  integer  but  for  all  values  of  m,  is  called  Legendre's  Equa- 
tion.    Its  solutions  are  discussed  in  Arts.  76,  85  and  86. 

74.  To  find  particular  solutions  of  the  equation 


which  is  equation  (11),  Art.  70,  when  z  is  independent  of  <£, 
let  z  =  R  -  T  where  R  is  a  function  of  r  alone  and  T  is  a  func- 
tion of  t  alone.  Substitute  in  the  equation. 


or 

JL    _ 

c*T  dt2  ~  R[dri+  r   dr  ] 

The  right-hand  member  of  (2)  does  not  involve  t.  Therefore 
the  left-hand  member  does  not.  The  left-hand  member  does  not 
involve  r.  Therefore  the  right-hand  member  does  not.  There- 
fore each  member  is  constant.  Call  the  constant  —  p?. 


(3) 

and 

dB 


sn      A 

+  -  -j-  +  p?K  =  0.  (4) 

^  r    dr  n 

A  particular  solution  of  (1)  is  therefore  R  •  T  where  T  is  de- 
termined by  equation  (3)  and  R  by  (4). 

Particular   solutions   of  equation    (3)    are    T  =  cos  pet   and 
T  =  smptt.      (See  Art.  31.) 

To  solve  equation   (4),    let  r  =  x/jj.  and   substitute  in    the 
equation. 

1   dR       _ 


Equation  (5)  is  a  special  case  of  the  more  general  equation 
affy±xdy±(a?       n^v       0 

>+      +  (   -n)y  =  v> 


INTEGRATION  IN  SERIES  103 

known  as  BesseFs  Equation.  Its  solutions  are  considered  in 
Arts.  79  to  84  inclusive. 

INTEGRATION  IN  SERIES 

75.  It  will  be  noted  that  as  yet  in  this  book  no  equations  with 
variable  coefficients,  of  higher  order  than  the  first,  have  been 
considered  except  a  few  very  special  cases  discussed  in  Chaps.  V 
and  VI.  The  remainder  of  this  chapter  is  devoted  to  a  discus- 
sion of  linear  differential  equations  of  the  second  order  with  coef- 
ficients rational  integral  functions  of  x,  and  second  member  zero. 
To  such  a  set  belong  Legendre's  and  Bessel's  Equations  men- 
tioned above. 

Not  all  differential  equations,  not  even  all  in  the  comparatively 
simple  form  of  linear  differential  equations  of  the  first  order,  are 
capable  of  solution  in  finite  form.  When  solutions  cannot  be 
found  in  finite  form,  recourse  is  had  to  integration  in  series.  In 
the  set  about  to  be  considered,  some  equations  have  solutions  in 
finite  form  and  some  have  not. 

We  shall  attempt  here  to  find  solutions  only  in  the  form  of 
infinite,  convergent,  power  series. 

If  an  equation  be  capable  of  solution  in  finite  form,  this  form 
is  found  when  a  solution  is  attempted  in  the  form  of  a  power  series. 
For  instance,  in  exercise  11,  page  120,  the  solution  found  as  if  it 
were  made  up  of  infinite  series  is  in  reality  in  finite  form. 

Sometimes  the  series  that  make  up  the  solution  of  an  equation 
may  be  recognized  as  those  of  familiar  functions.  In  such  cases, 
the  solution  can  be  written  in  terms  of  those  functions.  For 
instance,  in  the  answer  given  on  page  122  for  exercise  12,  page 
120,  if  A  be  taken  equal  to  2  and  B  to  1,  and  the  two  particular 
solutions  be  added,  there  results  the  series 

r-*  Fl  -L.  f^  -L.  (2a;)'  a.  (2*)'  -L  i.  (2*)n  j.         1 

*  L1+(2a:)  +  ir  IT       ~w     J' 

which  is  #~VZ.  If  the  second  solution  be  subtracted  from  the 
first,  there  results  the  series  which  is  x~*e~2x.  The  general  solu- 
tion is  therefore 


104    SHORT  COURSE  ON   DIFFERENTIAL  EQUATIONS 

y  =  c^-V  +  c2oTV2z. 
76.  Let  us  attempt  to  find  a  solution  of  Legendre's  Equation 


in  the  form  of  a  power  series  in  x. 

At  first,  assume  that  there  is  a  power  series 

y  =  g«*  +  9^+1  +  •  •  •  +  ^K+r  +  •  •  • 

v=Q 

where  g0,  gl9  •  •  •,  K  are  constants,  which  will  formally,  i.  e.., 
without  regard  to  whether  the  series  converges  or  not,  satisfy  the 
equation.  It  is  no  restriction  to  assume,  as  we  shall,  that  #0H=0, 
because,  if  there  is  any  solution  at  all,  one  at  least  of  the  t/'s  is 
not  zero,  and  we  assume  that  the  series  begins  with  the  term  con- 
taining the  first  g  which  does  not  vanish. 
Since 


and 

cW"^*"1" 

Substitute  in  the  equation. 


-1  +  m(m  +  1)^^+"]  =  0, 
or 


-{(K  +  v)(«  +  v  +  1)  _  m(m  +  1)}^«+-]  =  0. 
If 

y  =  S?^t+' 

i/=0 


INTEGRATION  IN  SERIES  105 

is  to  satisfy  the  equation,  the  coefficients  of  each  power  of  x  in 
(1)  must  be  zero.  Therefore  there  results  the  following  series 
of  equations  : 


1)  -  m(m  +  !)}&  =  0, 

2)  _  m(m  +  1)}^  =  0, 

•       '      '     (2) 


-  {(K  +  2r  -  2)(K  +  IT  -  1)  _  m(m 

From  the  first  of  these  equations,    since   gQ  =|=  0,   therefore, 
K  =  0,  or  K  =  1.     At  first,  take  K  =  1. 

Substitute  in  equation  (2)  and  calculate  the  #'s  in  succession. 

(m  —  l)(m  a.  2) 
&  =  0,     </2  =  -  ^        ->-       ^  (70,     ^3  =  0, 


(m+2r) 
~ 


where  ^0  is  arbitrary,  and  g2r  has  the  value  given  above,  formally 
satisfies  the  equation.  Since  this  series  is  convergent,  (3)  is  a 
particular  solution  of  the  equation. 

Next,  take  K  =  0. 

Substitute  in  equations  (2).     gl  is  arbitrary.     Call  it  zero. 


106    SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

m(m  -|-  1)  m(m_ 2)(m- 

ffi  =  ~         j~2         #o>  #3  =  ",  9*  =  ~  n; 

•••>  ^M=o, 

m(m-2)...(m-2r+2)(.. 
"ar  —  v  —  -1  y  1 2r 


y  - 


(4) 


where  g0  is  arbitrary,  and  #2r  has  the  value  given  above,  formally 
satisfies  the  equation.  Since  the  series  is  convergent,  (4)  is  a 
particular  solution  of  the  equation. 

If  solution  (3)  be  denoted  by  yl  and  solution  (4)  by  yv  the 
general  solution  of  the  equation  is  y  =  Ayv  -f  By^  where  A  and  B 
are  arbitrary  constants.  (See  Art.  11.) 

77.  The  general  form  of  a  linear  differential  equation  of  the 
second  order  with  right  hand  member  zero  is 

<?.(*)  gU  2^)1+  2,0*)  •</  =  <>.  (1) 

It  will  be  assumed  here  that  <?0(#),  ^C^),  ^2(^)  are  rational 
integral  functions  of  x. 

If  a  solution  is  to  be  found  in  the  form  of  a  power  series  in 
x  —  a,  it  will  be  convenient  to  write  the  equation  in  the  form 

(x  -  a)«p.(*)        +  (x  -  a)ft(aO  <     +  pt(x)  •  y  =  0,     (2) 


where  jt?0(#),  Pi(%),  P2(x)  are  rational  integral  functions  of  x. 
The  equation  can  be  written  in  this  form  in  an  unlimited  number 
of  ways  by  multiplying  it  through  by  a  suitable  power  of  x  —  a 
and  a  rational  fraction  neither  the  numerator  nor  denominator  of 
which  contains  x  —  a. 


INTEGRATION  IN  SERIES  107 

Definition.     The  point  a  is  a  regular  point  of  equation  (2)  if 

A(«)  4=  0. 

Without  at  first  making  any  assumption  with  regarol  to  the 
point  a,  substitute 

i/=0 

in  equation  (2)  and  attempt  to  determine  the  g's  so  that  the 

equation  is  formally  satisfied. 

i>=oo  r  "1 

v=o  [ 

=  0.      (3) 

Call  the  expression  in  square  brackets  /(#,  *  -f  v). 
Develop  f(x,  K  -f  v)  into  a  power  series  in  x  —  a  by  Taylor's 
Theorem. 

.-./(*,  K  +  V)  =/(a,  K  +  V)  +/'(a,  K  +  V)  -~-^+... 


12 

Substitute  this  development  in  (3),  equate   each  power  of 
x  —  a  to  zero  and  there  results  the  following  series  of  equations: 


+  2)  +  ^/'(a,  K  +  1)  +  ^0  =  0,       (4) 


f(n)(a   K) 
o^_>-l-^  =  0, 


=  0, 


Now  ^r0  =f=  0.     Therefore  /(a,  K)  =  0. 
And          /(a,  K)  =  K(K  _ 


108    SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 


2(a)  =0. 

From  this  equation  can  be  determined  the  value  or  values  of  K 
which  are  to  be  used  in  the  subsequent  equations  (4).     If  jo0(a) 
is  not  zero,  the  equation  is  of  the  second  degree.     If  J90(a)  is 
zero,  the  equation  is  of  lower  degree  than  the  second. 
The  necessary  and  sufficient  condition  that  the  equation 


is  of  the  second  degree  is  therefore  that  the  point  a  be  a  regular 
point  of  the  differential  equation. 

Definition.  The  equation  *(*  —  I)jo0(a)+K^i(a)  +  A(a)  =  0 
is  called  the  indicial  equation  of  the  differential  equation  (2). 

If  the  point  a  is  regular,  the  indicial  equation  gives  two  values 
of  *,  say  *'  and  *c",  and  from  equations  (4),  for  either  value  of 
*,  the  values  of  gv  g^  •  •  -,  may  be  computed,  in  general,  in 
terms  of  gQ. 

Therefore  in  general  there  are  two  series  in  ascending  powers 
of  x  —  a,  namely, 


and 


where  g0  is  arbitrary  in  either  series,  which  will  formally  satisfy 
equation  (2). 

78.  The  following  theorems  with  regard  to  the  solutions  of  the 
differential  equation 

O  -  «)W*)     +  O  -  °0^iO)     +  AGO  •  y  =  ° 


in  a  power  series  in  x  —  a  have  been  established.  The  proofs 
are  too  long  to  be  given  in  this  book.  For  a  discussion  of  these 
theorems  the  student  is  referred  to  a  pamphlet  entitled  '  '  Regu- 
lar Points  of  Linear  Differential  Equations  of  the  Second  Order  " 
by  Professor  Maxime  Bocher,  published  by  Harvard  University. 


INTEGRATION  IN  SERIES  109 

Theorem  I.  If  a  is  a  regular  point  of  the  differential  equa- 
tion, and  the  difference  of  the  roots  of  the  indicial  equation  is  not 
zero  or  a  positive  integer,  two  solutions  in  the  form  of  a  power 
series  in  x  —  a,  viz., 

y  =  I!  gv(x-a,y+v, 

V=Q 

and 


where  *'  and  K"  are  the  roots  of  the  indicial  equation,  exist,  and 
these  series  are  convergent.  In  each  of  these  series  gQ  is 
arbitrary. 

In  this  case,  if  yl  denotes  one  of  the  series  and  y2  the  other, 
the  general  solution  of  the  equation  is  y  =  Ayl  -j-  By2  where  A 
and  B  are  arbitrary  constants. 

A  case  to  which  this  theorem  applies  is  Bessel's  equation  when 
n  is  not  zero  nor  an  integer,  discussed  in  Art.  80. 

Note.  By  the  difference  of  the  roots  of  the  indicial  equation 
being  a  positive  integer  is  meant  that  the  greater  minus  the  less 
is  a  positive  integer. 

Theorem  II.  If  a  is  a  regular  point  of  the  equation  and  the 
difference  of  the  roots  of  the  indicial  equation  is  a  positive  inte- 
ger n,  the  necessary  and  sufficient  condition  that  two  solutions 
of  the  form  under  Theorem  I  exist  is  that 

f(n)(n    K"} 

?„-,/'(«.  *"  +  n  -  1)  +  •  •  •  +  sr/-1  ^  2  =  o, 

(see  equations  (4),  Art.  77),  where  K"  is  the  smaller  of  the 
roots,  and  when  this  condition  is  fulfilled,  the  series  are  conver- 
gent, 

In  this  case  the  series  corresponding  to  the  larger  value  of  K, 
say  K',  can  be  found  as  before.  In  the  series  corresponding  to 
K",  g0  and  gl  are  arbitrary,  but  if  gl  be  chosen  zero  a  particular 
solution  in  terms  of  g0  is  found.  Then  if  yl  denotes  the  first 
series,  and  yt  the  second,  the  general  solution  of  the  equation  is 
y  =  Ay^  +  By  -,  where  A  and  B  are  arbitrary  constants. 


110    SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

A  case  to  which  this  theorem  applies  is  Legendre's  Equation 
discussed  in  Art.  76.  , 

Theorem  III.  If  a  is  a  regular  point  of  the  equation,  and  the 
difference  of  the  roots  of  the  indicial  equation  is  either  zero,  or  a 
positive  integer  n  where 


two  solutions  are  found,  one  being 

y="f;</,O-«)"'+>', 

v=0 

the  other  being 

y  =  log  (x  _  a)  "if  gv(x  -  a)*'+*  +!!&(*  ~  a)*"+"» 

i/=0  v=0 

where  K'  is  the  larger  root  of  the  indicial  equation  and  K"  the 
smaller,  and  these  series  are  convergent.  In  the  first  of  these 
series  gQ  is  arbitrary.  In  the  second,  g0  is  arbitrary  and  gv  is 
determined  in  terms  of  g0. 

If  yl  denotes  the  first  series  and  yt  the  lasf  term  of  the  second, 
the  general  solution  of  the  equation  is 

y=[A  +  Blog(x-  a)]  yl  +  Byr 

where  A  and  B  are  arbitrary  constants. 

A  case  to  which  this  theorem  applies  is  Bessel's  Equation 
when  n  =  0,  or  an  integer,  discussed  in  Arts.  82  and  83. 

Theorem  IV.  If  the  point  a  is  not  a  regular  point  of  the 
equation  there  are  not  two  solutions  of  the  equation  in  any  of 
the  forms  under  Theorems  I  and  III,  and  if  any  series  in  one  of 
these  forms  is  found  it  is  usually  not  convergent. 

Cases  to  which  this  theorem  would  apply  will  not  be  considered 
in  this  book. 

BESSEL'S  EQUATION 

79.  We  shall  now  consider  the  solutions  of  the  equation 
o*^y  ±xdy  ±(tf      n^v      0 

M>+  dx  +  (      n)y  =  ()> 

in  the  form  of  a  power  series  in  x. 


INTEGRATION  IN  SERIES  111 

The  equation  as  it  stands  is  in  the  form  (2)  of  Art.  77,  where 

#>(*)  =  !>     -PiO)  =  *>     P*(?)  =  x*  -  n*' 
Since  j»0(#)  cannot  be  zero  for  any  value  of  x,  all  points  of 
this  equation  are  regular.     Therefore  the  solutions  of  the  equa- 
tion for  all  values  of  x,  and  in  particular  when  x  —  0,  will  come 
under  one  or  other  of  the  forms  mentioned  in  the  first  three 
theorems  of  Art.  78. 
Substitute 


v=Q 

in  the  equation. 


or 

"Z  [(K  +  v)2  +  a*  -  n^gvx*+»  =  0. 

v=Q 

The  equations  for  the  determination  of  the  </'s  are  therefore  : 

(«2  _  n*)g,  =  0, 

[(«  +  I)2  _  n^3l  =  0, 

[(*  +  2)2  -  WJ]s-2  +  gt  =  0, 

[(«  +  3)2  -  n2]^,  =  0,  (1) 

[(«  +  2r  _  I)2  -  n-]^,  =  0, 


Since  gr0  =(=  0,  from  the  first  equation  there  results  K  =  ±  n. 
The  difference  of  the  roots  of  the  indicial  equation  is  therefore 

±  2n.     If  n  =  0,  this  difference  is  zero.     If  n  =  ±  ^,  where  jo 

2i 

is  an  odd  integer,  or  if  n  is  an  integer,  this  difference  is  a  posi- 

f) 

tive  integer.     If  TI  is  neither  zero  nor  ±^  nor  an  integer,  the  dif- 

2i 

ference  is  neither  zero  nor  a  positive  integer. 

T) 

80.  At  first  assume  n  neither  zero  nor  ±  ^  nor  an  integer. 


112    SHORT  COURSE  ON  DIFFERENTIAL   EQUATIONS 

This  is  the  case  covered  by  Theorem  I.     There  are  therefore 
two  solutions 

y  =  "Z  9v*n+v 

v=0 

and 

y  =  "If  <7,*-n+". 

v=0 

To  determine  the  gv  substitute  K  =  n  in  equations  (1). 
•••0i  =  0,     &=  -  2(271  +  2)'     ^3=::0' 


'  4(2n  +  4)  ~  2  •  4(2n  +  2)(2w  +  4)  ' 


2  •  4  •  •  •  2r(2n  +  2)(2»  +  4)  •  •  •  (2w  +  2r) 


_i     _   _  ,    _j_  y^r**     I     .  /"O^ 

*"  2  -  4(2/1  +  2)(2n  +  4)  "  +  g,  * 

where  g0  is  arbitrary  and  g2r  has  the  value  given  above  is  a  par- 
ticular solution  of  the  equation. 

Similarly,   on  substituting  K  =  —  n  in  equations   (1)   there 
results 


2  •  4(2n  _  2)(2n_4) 
where  jr0  is  arbitrary  and 


___   __ 
y»  -  2  •  4  •  •  •  2r(2w  -  2)(2»  -  4)  •  •  •  (2»  -  2r)' 

is  a  particular  solution  of  the  equation. 


INTEGRATION  IN  SERIES  113 

If  yl  denotes  the  first  series  and  i/2  the  second,  the  general  solu- 
tion of  the  equation  is  y  =  Ayl  -}-  By2  where  A  and  B  are  arbi- 
trary constants. 

IY\ 

81.  Next,  assume  n  —  =t=~£.     Assume,  for  definiteness,  that  p 

2i 

is  positive.     In  this  case  the  difference  of  the  roots  of  the  indicial 
equation  is  a  positive  integer,  viz.  p.     From  an  examination  of 

f) 

equations  (1)  it  is  seen  when  n  =  —  ^  that  both  g0  and  gp  are 

2t 

arbitrary.     Choose  gp  =  0,  and  there  results  the  same  equation 

'P 
as  (3)  of  the  preceding  article  when  -f-  ^  is  substituted  for  n. 

Zt 

Therefore  in  this  case  there  are  two  particular  solutions  of  the 
equation  which  are  the  same  as  the  solutions  in  the  case  of  the 

f) 

preceding  article  when  ^  is  substituted  for  n. 
JL 

82.  Next,  assume  n  an  integer.     Since  n  appears  only  in  the 
form  of  a  square  in  the  differential  equation,  it  is  sufficient  to 
suppose  it  a  positive  integer. 

In  this  case  the  difference  of  the  roots  of  the  indicial  equation 
is  the  positive  integer  2n. 

For  the  root  *'  =  n,  the  series  is  the  same  as  (2)  of  Art.  80. 
For  the  root  K"  =  —  n,  the  equation 


is  such  that  the  coefficient  of  </2n  is  zero.  Therefore,  since 
#zn-2  4s  0>  tnis  case  comes  under  that  mentioned  in  Theorem  III, 
Art.  78. 

To  get  a  solution  corresponding  to  K",  let 


y  = 

i>=0  i>=0 

For  the  purpose  of  determining  the  coefficients  ~gv,  write  the 
series  in  the  form 

V  =  If  (<7,-2n  log  X  -f  ^,X'+", 


where  g_tn  =  g_,n+l  =  •  •  •  =  g^  =  0. 
9 


114    SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

Substitute  in  the  equation. 
•  •  -if  [(*"  +  ")2  +  xz  -  n^gv.^  log  x  -  **"+" 

"  nlgJxr+'^O.     (1) 


Now  in  the  coefficients  of  log  x  •  #*"+",  #_2n,  #_2n+1,  • 
are  zero,  and  the  remaining  ones  are  the  same  as  the  left  hand 
members  of  equations  (1),  Art.  79,  with  K"  -j-  2n  substituted  for 
K.  Now  K"  -j-  2n  =  *',  and  equations  (1),  Art.  79,  hold  for 
K  =  K'.  Therefore  all  the  coefficients  of  log  x  •  x«"+v  vanish. 

From  the  second  set  of  terms  in  (1)  are  found  the  equations 
from  which  to  determine  gv.     These  equations  are: 

(K"2   -  »')£.   =    0, 


[(""  +  2)'  -  <]?2  +  g0  =  0, 


in_,  +  2(<c"  +  2n)<70  =  0, 
I)2  -  n^g3n+l  +  2(K"  +  2n  +  1)^  =  0, 

=  0, 


Since  in  these  equations,  K"  =  —  n,  therefore  ^0,  an  arbitrary 
constant,  satisfies  the  first  equation.     Also, 


*2~  2(2n  -  2)' 

9o 

-4)' 


Since  the  coefficient  of  <?2n  is  zero,  this  equation  introduces  no 
new  g.  The  equation,  however,  gives  a  means  of  determining  the 
hitherto  undetermined  constant  g0, 


INTEGRATION  IN  SERIES 


115 


Also 


9**-*  = 


9o 


n     -   0       n    - «_        .  .  . 

9i-     >  01-2(2*1-2)' 
where 

So  =  -  22n 

Choose  </2n  =  0.     Therefore 


(2n  -  2)}2' 


n\n-lgQ. 


1         fl         1 


2-4 


I1  , '      !     1 
J2+2nZ2P«      ' 

I  — i        All— }—  Z/  I 


.  •  .  y  = 


+"  -  22n~1  In  In- 


{2.4-6. 


rn+ 


r  -j  -j 

-J-_L 

2)  12  +  2^  + 


-  2) 


rrn+2r  J_  . 

2r 


where  gQ  is  arbitrary  and  ^2n+2r  has  the  value  given  above,  is  a 
particular  solution  of  the  equation 

If  the  first  solution  be  denoted  by  ^  and  all  terms  not  involv- 
ing log  x  in  the  second  by  yv  the  general  solution  of  the  equation 
is  y  —  ( A  -f-  B  log  x)yl  -f-  By^  where  A  and  B  are  arbitrary 
constants. 


116    SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

83.  Next,  assume  n  =  0.  In  this  case  the  difference  of  the 
roots  of  the  indicial  equation  is  zero.  This  case  is  covered  by 
Theorem  III. 

The  two  infinite  series  when  n  =  0  can  be  found  from  equa- 
tions derived  as  in  the  preceding  case.  They  can  also  be  found 
by  letting  n  be  zero  in  the  results  in  that  case.  Two  particular 
solutions  are 

~2  r4  r2r 

x        ...JL.C    iv  _  -  ___ 

t~22-42~~  "^       ;  22-  42-  •  •  (2r)2 

and 


84.  As  will  appear  in  applications  to  physical  problems,  when 
n  is  a  positive  integer  it  is  convenient  to  take,  not  yl  and  yv  but 
the  quotients  of  these  by  2"  n  where  g0  is  unity.  These  special 
solutions  are  written  Jn(x)  and  TFn(#)  so  that 


2)(2n  +  4) 


and 

(2»-3 1  n  __  2  x~ 
2»-l\n-lx-»+- 


2n~ln-i 


LEGENDRE'S  EQUATION 
85.  Returning  now  to  the  equation 


considered  in  Art.  76,  we  see  that  it  can  be  transformed  into  the 


INTEGRATION  IN  SEEIES  117 

form  (2)  of  Art.  77  by  multiplying  through  by  x*.     The  trans- 
formed equation  is 

*')        -  2*°       +  m(m  +  l)**y  =  0, 


where  pQ(x)  =  1  —  x2,  jf?,(^)  =  —  2#2,  ^>2(#)  =  m(m  - 
Since  pQ(x)  =  0  when  x  =  —  1  and  #  =  1,  the  points  —  1  and 
1  are  not  regular  points  of  the  equation.  All  other  points  are 
regular. 

The  indicial  equation  when  x  =  0  is 

K(K-  1)  =0. 

This  equation  has  the  roots  0  and  1.  The  differential  equa- 
tion in  this  case  comes  under  the  case  mentioned  in  Theorem  II. 
This  case  was  already  discussed  in  Art.  76. 

86.  It  is  convenient  when  m  is  a  positive  integer  as  it  was  in 
the  illustration  of  Art.  73  to  have  a  solution  of  Legendre's  Equa- 
tion as  a  series  in  descending  powers  of  x.  In  this  case  we  shall 
take  the  equation,  as  in  Art.  73,  in  the  form 


Let 


and  substitute  in  the  equation. 
.•.*£  (n-vXn-v-l)?,*—  "* 

v=0 

_  [(n  —  v)(n  —  v  4-  1)  —  m(m  +  l^g^-*  =  0. 

.  •  .  [n(n  +  1)  _  m(m  +  I)]0r0  =  0, 

[O  _  l)n  _  m(m  +  1)]^  =  0, 

w(w  -  IX  -  [(^  -  2;(n  -  1)  -  m(m  +  1)]^  =  0, 


-  [(n  -  2r)(n  -  2r  +  1)  _  m(m  +  l)]^r  =  0. 


118    SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

From  the  first  equation,  since  g0  =4=  0> 

.  •  .  n(n  +  1)  —  m(m  -f  1)  =  0. 
.-.n  =  m     or     n  =  —  m  —  1. 
"it  first  take  n  =  m. 

(m  —  l)m 
~ 


=  -  2  (2m  -~T) 

(m-2r-3)(m-2r-2)...(m-l  )m 
to--»AT-'-^     2r  •  •  •  4  •  2(2m  _  2r  +  1)  •  •  •  (2m  _  1)    9o' 

A  series  in  descending  powers  of  x  which  satisfies  the  equation 
is  therefore 


^  1 

h  ^  J    ( 


(m  —  3)(m  —  2)(m  —  l)m 
+     4.2(2m_3)(2m_  iy 

lere  g0  is  arbitrary  and  #2r  has  the  value  given  above. 
By  taking  n  =  —  m  —  1,  there  results  the  solution 

(m  -|-  2)(m 


2(2m  +  3) 


(m  +  4)(m  +  3)(m  +  2)(m  +  1)  jr., 


where  ^0  is  arbitrary  and 

(m  +  2r)(m  +  2r  -  1)  •  •  •  (m  +  1) 
^2r  ~  2r  •  •  •  2(2m  +  3)  •  •  •  (2m  +  2r  +  1)  ^ 

When  m  is  a  positive  integer,  solutions  (3)  and  (4)  of  Art. 
76  is  a  finite  series  according  as  m  is  even  or  odd,  and  in  either 
case,  equation  (1)  above  is  finite  differing  from  (3)  or  (4)  of 
Art.  76  only  by  a  constant  factor. 

87.    If     series      (1)      of     Art.     86      be      multiplied      by 
-3)...l   Qr  _^m_  ^  resulti       .          ^  ig 
2m(jm)2 


INTEGRATION  IN  SERIES  119 

called  the  Legendrian  Coefficient  of  the  mth  order,  and  is  de- 
noted by  Pm(x). 

The  successive  values  of  Pm(x)  are  readily  found  to  be 


EXERCISES 

1.  Find  the  remaining  six  particular  solutions  of  the  equation 
considered  in  Art.  72. 

2.  Find  two  particular  solutions  of  the  equation 

•  du        2d2u 

at  =  c  ~dtf  * 

3.  Find  two  particular  solutions  of  the  equation 

d\ru± 


dt  dr2 

4.  Find  four  particular  solutions  of  the  equation 
du 


_   t 

=    *  \da?  +  d 

5.   Find  four  particular  solutions  of  the  equation 


_ 

at2  ~     dtf  ' 

6.   Find  four  particular  solutions  of  the  equation 


1.  Show  that  equation  (1)  of  Art.  70,  in  rectangular  coor- 
dinates, becomes  equation  (2)  when  transformed  to  polar  or 
spherical  coordinates.  The  equations  of  transformation  are 
x  =  r  cos  0  sin  <£,  y  =  r  sin  0  sin  <£,  z  =  r  cos  <£. 

du      du  dr       du  dO      du  d<t>          »    •    si    i 
Suggestion.     ^-  =  ^-  *-  +  ^  jr  +  ^  tr>  and  similarly 
dx       dr  dx  ^  dO  dx  '   d<f>  ox 

for  y  and  & 


120    SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 
Find  the  general  solutions  of  the  following  equations  : 


9.  9^        +  (9*  -  a?)       -  (*  +  l)y  =  0. 


11.  A-  2^ 


14.   (*- 


17.  Show  that, 

J.'(y)  =  -  J,(y), 


ANSWERS 

1.  z  =  sin  cue  cos  /3y  sin  d 
2  =  sin  cu;  cos  ^?/  cos  c£ 

2  =  cos  ox  sin  /fy  sin  ct  Va2  +  y82, 
z  =  cos  cue  sin  /3y  cos  c£  V«2  +  )82, 
2  =  cos  cue  cos  fiy  sin  c^  Va2  +  P2, 

Z  =  COS  OX  COS  /ty  COS  d  Va2  -j-  /32. 

2.  16  =  e  ~c2ft2'  cos  cur,     ^  =  e~c2a2i  sin  cue. 


INTEGRATION  IN  SERIES  121 

3.  u  =  -  e-°2a2t  cos  ar,     u  =  -  e~c2a2t  sin  ar. 

r  r 

4.  u  =  e~c2(a2+^2x  cos  a#  Cos  Py,     u  =  6-c2(a2+£2)'  cos  ax  sin  y^ 
^  =  e-<2(«2+02)<  sin  cue  sin  /3y,      u  =  e-'2(*2+0 

5.  i/  =  cos  ax  cos  ca£,     y  =  sin  aa;  sin  ca£, 
2/  =  sin  cu;  cos  cat,     y  =  cos  ax  sin  ca£. 


6.  i/  =  e  kt  sin  aa;  cos  t  Vc2a2  —  ^2,     y  =  e  kt  sin  cue  sin  <  VcV  _  k2, 
y  —  e'™  cos  ax  cos  £  VcV  _  k2,     y  =  e"*'  cos  cur  sin  £  Vc2a2  —  k2. 


where 
and 


2r  =     "        2-4...2r.3.7...(4r_l)' 
9. 

i  if-,        4  4-7  4-740      , 

y  =  A*    1  +  3^-^+  3^6^+3^546:33"  +' 


where 


and     0  =  =- 


~  32*-  •  5  •  •  •  (2r  +  3r2)  ~  32'  • 1  •  •  •  (3r2  _  2r) 

10.  y  =  Aa£    1  +  g^  +  g^Tgrj-Q  +  ' '  *  +  &Xr  H-  •  V 


122    SHORT  COURSE  ON  DIFFERENTIAL  EQUATIONS 

where 

__  1  1 

9"~  8'-3-10---(2r2  +  r)  9tr~  8' 

11.  y  =  An?  +  -Br. 

12.  =  J* 


+  Jfcr2  [l  +  j*°  +  ^2  *•  +  •  •  •  +  g^f  +  •  •  •  J  , 

where 

_  $_  _  4r 

^  =  6-20---2(2r»+r)  ^  =  2  -  12  •  •  •  2(2r2-  r)' 

13.  y  =  ^  1^1  +  TL^  +  j^^^6  +  '  '  •  +9^'  +  •  •  -J 


where 

1 _         1_ 

9*r  ~  12  -  42  -  •  -  SCSr2  +  r)  ^Sr  ~  6  •  30  •  •  •  3(3r>  -  rj' 

r         2  22      2  23  1 

[22            23  24 

/r              rvi  _j_  IA  I              7-3n  _j_ !  _i_n 

ir*  —  ijl    i>2" ^  v -1  +  ^"/  +  ^2    02    02 ^  L1  H~  ~Z  ~r  3 J 

J.                  J.     •  L  1     *  A    '  O 

25  1 

- 1^2273274-2  ^L1  +  4  +  4  + 1]  +  •••  +  yX  +  •••[! 

where 

or  Qr+l  r=r  1 

^r=  (-  I)r12>2t     v     and     ^r  =  (-  l)r12T|2T77?r2Z  -- 


INTEGRATION  IN  SERIES  123 

16.  y= 


where 


2r  =  2'-42---(2r)»rl  r 


»  ' 


FISHER  AND  SCHWATPS 

Series  of  Text-Books  on  Mathematics 

Rudiments  of  Algebra $  .60  School  Algebra $J*00 

Secondary  Algebra J.08  Elements  of  Algebra JJO 

Complete  Secondary  Algebra. .  \  .35  Higher  Algebra \ «50 

Quadratics  and  Beyond 90  Text  Book  of  Algebra,  Part  I.  MO 


TIHE  Fisher  and  Schwatt  algebras  have  achieved  a  marked  success 
I  in  the  short  time  that  they  have  been  before  the  public.  The 
"  Higher  Algebra"  is  used  as  a  regular  text-book  at  Harvard 
University,  and  the  elementary  books  are  established  in  sec- 
ondary schools  of  corresponding  rank. 

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which  consists  of  the  "Rudiments"  (for  grammar  schools)  and  the  "Sec- 
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Summation  of  Series,  the  Exponential  and  Logarithmic  Series,  Determin- 
ants, and  Theory  of  Equations.  Thus,  while  the  "Secondary"  covers 
the  requirements  for  admission  to  colleges,  the  "Complete  Secondary" 
covers  the  requirements  for  admission  to  any  scientific  school,  and  is  also 
sufficiently  full  for  the  ordinary  work  of  the  first  year  in  college.  "  Quad- 
ratics and  Beyond"  consists  of  the  second  half  of  the  "Complete  Sec- 
ondary," bound  separately  for  the  convenience  of  advanced  and  reviewing 
classes  in  secondary  schools,  and  for  college  freshman  work. 

The  second  series  constitutes  a  more  difficult  course  and  places  more 
emphasis  on  the  theory  of  mathematics.  In  this  series  the  "School 
Algebra"  corresponds  roughly  to  the  "Secondary,"  and  the  "Elements" 
to  the  "  Complete  Secondary." 

The  "Higher  Algebra "  is  a  book  for  college  courses.  The  "Textbook 
of  Algebra"  is  a  high  school  book  containing  an  unusually  large  number 
of  graded  exercises ;  it  is  a  valuable  mine  of  problems  bearing  on  the 
regular  high  school  work. 


THE   MACMILLAN    COMPANY 

64-66  FIFTH  AVE..  NEW  YORK 
BOSTON  CHICAGO  SAN  FRANCISCO  ATLANTA 


The  Elements  of  the  Differential 
and  Integral  Calculus 

By  DONALD  FRANCIS  CAMPBELL 

Professor  of  Mathematics  in  Armour  Institute  of  Technology 

i2mo.     Cloth,    x -[-364  pages.    $1.90  net 

•HIS  book  is  designed  especially  to  introduce  the  student 
of  engineering  to  the  mathematics  upon  which  his  future 
work  will  be  based,  but  in  spite  of  the  emphasis  on  the 
practical  side  of  the  subject  the  needs  of  classes  in  clas- 
sical colleges  and  universities  have  been  neglected  in  no  way. 

To  meet  the  needs  of  colleges  where  the  study  of  the  calculus 
is  taken  up  in  the  first  year  of  the  course,  Professor  Campbell  has 
presented  a  more  detailed  discussion  in  the  opening  of  both  the 
differential  and  integral  parts  of  his  work  than  is  usually  given 
in  text-books.  Thereafter,  the  subject  is  developed  by  the  use  of 
practical  problems  which  are  sure  to  arise  in  engineering  work. 
Thus  all  subjects  only  remotely  connected  with  engineering  have 
been  omitted,  while  in  addition,  a  few  elementary  chapters  in 
mechanics  have  been  supplemented.  This  presentation  of  mate- 
rial, without  encumbering  the  book,  affords  a  short  introduction 
to  Mechanics  and  Differential  Equations  as  well  as  a  view  of  the 
principles  of  Attraction,  Centers  of  Gravity,  and,  to  a  certain  ex- 
tent, the  Moments  of  Inertia,  from  the  mechanical  rather  than 
from  the  purely  mathematical  side. 

The  part  of  the  book  which  differs  most  widely  from  other 
text-books  is  that  dealing  with  the  integral  calculus.  A  full  ex- 
planation is  given  of  each  step  in  the  formation  of  each  summation 
and  integral.  In  addition,  in  order  to  enable  the  student  to  grasp 
more  fully  the  details  of  the  subject,  the  author  has  introduced  a 
large  number  of  practical  questions  which  are  found  in  actual  ex- 
perience to  produce  the  desired  result  better  than  the  theoretical 
propositions  introduced  into  the  older  treatises. 


THE  MACMILLAN  COMPANY 

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BOSTON  CHICAGO  SAN  FRANCISCO  ATLANTA 


14  DAY  USE 

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4IAR.18.1964 


LD  21-50m-12,'61 
(C4796slO)476 


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